Proof step 1 from proof inv-get-inverse-between (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  nσ(inv(nσ)(n) + m + 1) ≠ n + 1, inv(nσ)(n) + m + 1 < inv(nσ)(n + 1), ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ) ⊦

Rule argument

Additional information

Proof step 2 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  nσ(inv(nσ)(n) + m + 1) ≠ n + 1, inv(nσ)(n) + m + 1 < inv(nσ)(n + 1), ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 3 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ stutf(nσ) → inv(nσ) = (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ to leninv(nσ).n)

• Proof step 5:

  nσ ( (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ to leninv(nσ).n)(n) + m + 1) ≠ n + 1, (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ to leninv(nσ).n)(n) + m + 1 < (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ to leninv(nσ).n)(n + 1), ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ) ⊦

Rule argument

Additional information

Used terms: nσ

Proof step 4 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma inv-def

Proof step 5 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  nσ ( (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to leninv(nσ).n⌝ )(n) + m + 1) ≠ n + 1, (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to leninv(nσ).n⌝ )(n) + m + 1 < (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to leninv(nσ).n⌝ ) (n + 1), ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ⌜nf⌝ to n = ⌜nf to n⌝ ; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic

Proof step 6 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 7:

  leninv(nσ) = ∞, nσ ( (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to leninv(nσ).n⌝ )(n) + m + 1) ≠ n + 1, ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ (n) + m + 1 < (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to leninv(nσ).n⌝ ) (n + 1), ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ) ⊦

• Proof step 80:

  leninv(nσ) ≠ ∞, nσ ( (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to leninv(nσ).n⌝ )(n) + m + 1) ≠ n + 1, ⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to leninv(nσ).n⌝ (n) + m + 1 < (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to leninv(nσ).n⌝ ) (n + 1), ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ) ⊦

Rule argument

Additional information

Proof step 7 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  # nσ = ∞, nσ((leninv(nσ) = ∞ ⊃ ⌜λ m0. max∞(λ n0. nσ(n0) = m0).n⌝ ;⌜(λ m0. max∞(λ n0. nσ(n0) = m0).n) to leninv(nσ).n⌝ )(n) + m + 1) ≠ n + 1, max∞(λ n0. nσ(n0) = n).n + m + 1 < (leninv(nσ) = ∞ ⊃ ⌜λ m0. max∞(λ n0. nσ(n0) = m0).n⌝ ;⌜(λ m0. max∞(λ n0. nσ(n0) = m0).n) to leninv(nσ).n⌝ )(n + 1), ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0) ⊦

Rule argument

Additional information

Used simplifier rules: leninv-inf-strong: ⊦ stutf(nσ) → ( leninv(nσ) = ∞ ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)) ∧ (∀ m. ∃ m0. m < nσ(m0))); ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ⌜m⌝ < ∞; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜nf⌝ (n) = nf(n); (s)

Proof step 8 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

  ⊦ stutf(nσ) → (leninv(nσ) = ∞ ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)))

• Proof step 10:

  # nσ = ∞, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), max∞(λ n0. nσ(n0) = n).n + m + 1 < (# nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)) ⊃ ⌜λ m0. max∞(λ n0. nσ(n0) = m0).n⌝ ;⌜(λ m0. max∞(λ n0. nσ(n0) = m0).n) to leninv(nσ).n⌝ ) (n + 1) ⊦ nσ ( (# nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)) ⊃ ⌜λ m0. max∞(λ n0. nσ(n0) = m0).n⌝ ;⌜(λ m0. max∞(λ n0. nσ(n0) = m0).n) to leninv(nσ).n⌝ ) (n) + m + 1) = n + 1

Rule argument

Additional information

Used terms: nσ

Proof step 9 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma leninv-inf

Proof step 10 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  # nσ = ∞, nσ(max∞(λ n0. nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. nσ(n0) = n).n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜nf⌝ (n) = nf(n); (s)

Proof step 11 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 12:

  # nσ = ∞, nσ(max∞(λ n0. nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. nσ(n0) = n).n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

• Proof step 67:

  # nσ = ∞, nσ(max∞(λ n0. nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. nσ(n0) = n).n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0) ⊦ ∀ n. ∃ m. nσ(m) = n

Rule argument

Additional information

Used formulas: ∀ n. ∃ m. nσ(m) = n

Proof step 12 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 13:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ∞, nσ(max∞(λ n0. nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. nσ(n0) = n).n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

• Proof step 31:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) ≠ ∞, nσ(max∞(λ n0. nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. nσ(n0) = n).n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used formulas: max∞(λ n0. nσ(n0) = n) = ∞

Proof step 13 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ∞, nσ(∞.n + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Proof step 14 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 15:

  ⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))

• Proof step 16:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n. ∃ m. nσ(m) = n, (∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = n) ∨ (∀ n0. nσ(n0) ≠ n) ⊦

Rule argument

Additional information

Used terms: λ n0. nσ(n0) = n

Proof step 15 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma maxinf-inf

Proof step 16 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 17:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), ∀ n. ∃ m. nσ(m) = n, ∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = n ⊦

• Proof step 28:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), ∀ n. ∃ m. nσ(m) = n, ∀ n0. nσ(n0) ≠ n ⊦

Rule argument

Additional information

Proof step 17 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 18:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∃ m. nσ(m) = n, ∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = n ⊦

Rule argument

Additional information

Used terms: n

Proof step 18 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ nσ(m0) + 1, ⌜nσ(m0)⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m0) + 1).n, ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = nσ(m0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 19 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 20:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ nσ(m0) + 1, ⌜nσ(m0)⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m0) + 1).n, ∀ m. ∃ m0. m < nσ(m0), ∃ m1. m0 < m1 ∧ nσ(m0) < nσ(m1), ∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = nσ(m0) ⊦

Rule argument

Additional information

Used terms: m0

Proof step 20 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ nσ(m0) + 1, m0 < m1, ⌜nσ(m0)⌝ + ⌜1⌝ < leninv(nσ), nσ(m0) < nσ(m1), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m0) + 1).n, ∀ m. ∃ m0. m < nσ(m0), ∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = nσ(m0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 21 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 22:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ nσ(m0) + 1, m0 < m1, nσ(m0) < nσ(m1), ⌜nσ(m0)⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m0) + 1).n, ∀ m. ∃ m0. m < nσ(m0), ∃ n1. m1 < n1 ∧ nσ(n1) = nσ(m0) ⊦

Rule argument

Additional information

Used terms: m1

Proof step 22 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  # nσ = ∞, nσ(n1) = nσ(m0), nσ(∞.n + m + 1) ≠ nσ(m0) + 1, m0 < n1, m0 < m1, m0 + 1 < n1, m1 < n1, nσ(m0) < nσ(m1), ⌜nσ(m0)⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m0) + 1).n, ∀ m. ∃ m0. m < nσ(m0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m.+1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic trans-Nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m.+1 < n0; (forward) from specification nat-basic

Proof step 23 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 24:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

• Proof step 25:

  # nσ = ∞, nσ(n1) = nσ(m0), nσ(∞.n + m + 1) ≠ nσ(m0) + 1, m0 + 1 < n1, m0 < m1, m0 < n1, m1 < n1, ⌜nσ(m0)⌝ + ⌜1⌝ < leninv(nσ), nσ(m0) < nσ(m1), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m0) + 1).n, ∀ m. ∃ m0. m < nσ(m0), mon≤(nσ) ∧ m1 ≤ m0 ∧ ⌜m0⌝ < # nσ → nσ(m1) ≤ nσ(m0) ⊦

Rule argument

Additional information

Used terms: nσ, m1, m0

Proof step 24 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 25 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  # nσ = ∞, nσ(n1) = nσ(m0), nσ(∞.n + m + 1) ≠ nσ(m0) + 1, m0 < n1, m0 < m1, m0 + 1 < n1, m1 < n1, nσ(m0) < nσ(m1), ⌜nσ(m0)⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m0) + 1).n, ∀ m. ∃ m0. m < nσ(m0) ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → (n ≤ m ↔ false); (s)

Proof step 26 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 27:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → m ≤ n; (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: nσ, m1, n1

Proof step 27 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 28 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 29:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∃ m. nσ(m) = n, ∀ n0. nσ(n0) ≠ n ⊦

Rule argument

Additional information

Used terms: n

Proof step 29 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ nσ(m0) + 1, ⌜nσ(m0)⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m0) + 1).n, ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n0. nσ(n0) ≠ nσ(m0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 30 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m0

Proof step 31 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 32:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , max∞(λ n0. nσ(n0) = n) ≠ ∞, nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. nσ(n0) = n), n1

Proof step 32 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 33 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 34:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , max∞(λ n0. nσ(n0) = n + 1) = ∞, nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

• Proof step 47:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , max∞(λ n0. nσ(n0) = n + 1) ≠ ∞, nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used formulas: max∞(λ n0. nσ(n0) = n + 1) = ∞

Proof step 34 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 35:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , max∞(λ n0. nσ(n0) = n + 1) = ∞, nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < ∞.n, stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 35 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 36:

  ⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))

• Proof step 37:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < ∞.n, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), ∀ n. ∃ m. nσ(m) = n, (∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = n + 1) ∨ (∀ n0. nσ(n0) ≠ n + 1) ⊦

Rule argument

Additional information

Used terms: λ n0. nσ(n0) = n + 1

Proof step 36 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma maxinf-inf

Proof step 37 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 38:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < ∞.n, stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∃ m. nσ(m) = n + 1, (∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = n + 1) ∨ (∀ n0. nσ(n0) ≠ n + 1) ⊦

Rule argument

Additional information

Used terms: n + 1

Proof step 38 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , nσ(m0) = n + 1, nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < ∞.n, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), (∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = n + 1) ∨ (∀ n0. nσ(n0) ≠ n + 1) ⊦

Rule argument

Additional information

Proof step 39 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 40:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , nσ(m0) = n + 1, nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < ∞.n, stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = n + 1 ⊦

• Proof step 46:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , nσ(m0) = n + 1, nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < ∞.n, stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n0. nσ(n0) ≠ n + 1 ⊦

Rule argument

Additional information

Proof step 40 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 41:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , nσ(m0) = n + 1, nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < ∞.n, stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∃ m1. m0 < m1 ∧ nσ(m0) < nσ(m1), ∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = n + 1 ⊦

Rule argument

Additional information

Used terms: m0

Proof step 41 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 42:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , nσ(m0) = n + 1, nσ(n1 + m + 1) ≠ n + 1, m0 < m1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n + 1 < nσ(m1), n1 + m + 1 < ∞.n, stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = n + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic

Proof step 42 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 43:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , nσ(m0) = n + 1, nσ(n1 + m + 1) ≠ n + 1, m0 < m1, n + 1 < nσ(m1), ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < ∞.n, stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∃ n1. m1 < n1 ∧ nσ(n1) = n + 1 ⊦

Rule argument

Additional information

Used terms: m1

Proof step 43 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , nσ(m0) = n + 1, nσ(n0) = n + 1, nσ(n1 + m + 1) ≠ n + 1, m0 < n0, m0 < m1, m0 + 1 < n0, m1 < n0, n + 1 < nσ(m1), ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < ∞.n, stutf(nσ), ∀ m. ∃ m0. m < nσ(m0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m.+1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic trans-Nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m.+1 < n0; (forward) from specification nat-basic

Proof step 44 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 45:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → m ≤ n; (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: nσ, m1, n0

Proof step 45 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 46 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m0

Proof step 47 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 48:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , max∞(λ n0. nσ(n0) = n + 1) = ⌜n2⌝ , max∞(λ n0. nσ(n0) = n + 1) ≠ ∞, nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < n2, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. nσ(n0) = n + 1), n2

Proof step 48 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 49:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , max∞(λ n0. nσ(n0) = n + 1) = ⌜n2⌝ , nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < n2, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 49 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 50:

  ⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)

• Proof step 51:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , nσ(n2) = n + 1, nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < n2, stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. nσ(m) = n + 1 → ¬ n2 < m, ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used terms: n2, λ n0. nσ(n0) = n + 1

Proof step 50 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma maxinf-fin

Proof step 51 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 52:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , nσ(n2) = n + 1, nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < n2, stutf(nσ), ∀ m. nσ(m) = n + 1 → ¬ n2 < m, ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Proof step 52 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 53:

  ⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)

• Proof step 54:

  # nσ = ∞, nσ(n2) = n + 1, nσ(n1) = n, nσ(n1 + m + 1) ≠ n + 1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < n2, stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. nσ(m) = n + 1 → ¬ n2 < m, ∀ m. nσ(m) = n → ¬ n1 < m, ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used terms: n1, λ n0. nσ(n0) = n

Proof step 53 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma maxinf-fin

Proof step 54 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 55:

  # nσ = ∞, nσ(n2) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n2), n1 + m + 1 < n2, ⌜nσ(n1)⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. nσ(m) = nσ(n2) → ¬ n2 < m, ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Proof step 55 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 56:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

• Proof step 57:

  # nσ = ∞, nσ(n2) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n2), n1 + m + 1 < n2, ⌜nσ(n1)⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. nσ(m) = nσ(n2) → ¬ n2 < m, ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, mon≤(nσ) ∧ n2.-1 ≤ n2 ∧ ⌜n2⌝ < # nσ → nσ(n2.-1) ≤ nσ(n2), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used terms: nσ, n2.-1, n2

Proof step 56 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 57 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 58:

  # nσ = ∞, nσ(n2) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n2), n1 + m + 1 < n2, ⌜nσ(n1)⌝ + ⌜1⌝ < leninv(nσ), nσ(n2 - 1) ≤ nσ(n2), stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. nσ(m) = nσ(n2) → ¬ n2 < m, ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ ⌜m⌝ < ∞; (s) ⊦ n.-1 = n - 1; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m ≤ m + n; (s) n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0; (s) ⊦ m < n → n ≠ 0; (s) ⊦ m < 1 ↔ m = 0; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss)

Proof step 58 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 59:

  # nσ = ∞, n2 = m1 + 1, nσ(n2) = nσ(n1) + 1, n2 ≠ 0, nσ(n1 + m + 1) ≠ nσ(n2), n1 + m + 1 < n2, nσ(m1) ≤ nσ(n2), ⌜nσ(n1)⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. nσ(m) = nσ(n2) → ¬ n2 < m, ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used simplifier rules: stutf-zero: ⊦ stutf(nσ) → nσ(0) = 0; ⊦ m + 1 ≠ 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n2, m1

Proof step 59 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 60:

  # nσ = ∞, nσ(m1 + 1) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m < m1, nσ(m1) ≤ nσ(n1) + 1, ⌜nσ(n1)⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. nσ(m) = nσ(n1) + 1 → ¬ m1 + 1 < m, ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m + 1 ≠ 0; (s)

Proof step 60 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 61:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

• Proof step 62:

  # nσ = ∞, nσ(m1 + 1) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m < m1, ⌜nσ(n1)⌝ + ⌜1⌝ < leninv(nσ), nσ(m1) ≤ nσ(n1) + 1, stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. nσ(m) = nσ(n1) + 1 → ¬ m1 + 1 < m, ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, mon≤(nσ) ∧ n1 ≤ n1 + m + 1 ∧ ⌜n1 + m + 1⌝ < # nσ → nσ(n1) ≤ nσ(n1 + m + 1), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used terms: nσ, n1, n1 + m + 1

Proof step 61 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 62 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 63:

  # nσ = ∞, nσ(m1 + 1) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m < m1, ⌜nσ(n1)⌝ + ⌜1⌝ < leninv(nσ), nσ(m1) ≤ nσ(n1) + 1, nσ(n1) ≤ nσ(n1 + m + 1), stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. nσ(m) = nσ(n1) + 1 → ¬ m1 + 1 < m, ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ m ≤ m + n; (s)

Proof step 63 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 64:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

• Proof step 65:

  # nσ = ∞, nσ(m1 + 1) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m < m1, nσ(n1) ≤ nσ(n1 + m + 1), nσ(m1) ≤ nσ(n1) + 1, ⌜nσ(n1)⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. nσ(m) = nσ(n1) + 1 → ¬ m1 + 1 < m, ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), mon≤(nσ) ∧ n1 + m + 1 ≤ m1 ∧ ⌜m1⌝ < # nσ → nσ(n1 + m + 1) ≤ nσ(m1), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used terms: nσ, n1 + m + 1, m1

Proof step 64 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 65 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 66:

  # nσ = ∞, nσ(n1) = nσ(n1 + m + 1), nσ(m1 + 1) = nσ(n1) + 1, n1 + m < m1, nσ(m1) ≤ nσ(n1) + 1, ⌜nσ(n1)⌝ + ⌜1⌝ < leninv(nσ), nσ(n1) ≤ nσ(m1), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. nσ(m) = nσ(n1) + 1 → ¬ m1 + 1 < m, ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m. ∃ m0. m < nσ(m0), ∀ n. ∃ m. nσ(m) = n ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m ≤ m; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m ≤ m + n; (s) ⊦ m = m + n ↔ n = 0; (s) let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ m ≤ n → (n ≤ m ↔ m = n); (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ m + 1 ≤ n ↔ m < n; (s)

Proof step 66 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: n1 + m + 1

Proof step 67 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 68:

  # nσ = ∞, nσ(max∞(λ n0. nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. nσ(n0) = n).n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. ∃ m0. m < nσ(m0), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0) ⊦ ∃ m. nσ(m) = n0

Rule argument

Additional information

Proof step 68 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule weakening

And obtained the following new goals

• Proof step 69:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0) ⊦ ∃ m. nσ(m) = n0

Rule argument

Additional information

Proof step 69 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 70:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0) ⊦ ∃ m. nσ(m) = 0

• Proof step 71:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ nσ. # nσ = ∞ ∧ stutf(nσ) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)) → (∃ m. nσ(m) = n) ⊦ ∃ m. nσ(m) = n.+1

Rule argument

Additional information

Used simplifier rules: Nat freely generated by 0, +1;

Proof step 70 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: stutf-zero: ⊦ stutf(nσ) → nσ(0) = 0; Used terms: 0

Proof step 71 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 72:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ nσ. # nσ = ∞ ∧ stutf(nσ) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)) → (∃ m. nσ(m) = n) ⊦ ∃ m. nσ(m) = n + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m.+1 = m + 1; (s)

Proof step 72 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 73:

  # nσ = ∞, stutf(nσ), # nσ = ∞ ∧ stutf(nσ) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)) → (∃ m. nσ(m) = n), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0) ⊦ ∃ m. nσ(m) = n + 1

Rule argument

Additional information

Used terms: nσ

Proof step 73 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 74:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0) ⊦ ∃ m. nσ(m) = nσ(n0) + 1

Rule argument

Additional information

Proof step 74 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 75:

  # nσ = ∞, stutf(nσ), ∃ m0. n0 < m0 ∧ nσ(n0) < nσ(m0) ⊦ ∃ m. nσ(m) = nσ(n0) + 1

Rule argument

Additional information

Used terms: n0

Proof step 75 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 76:

  # nσ = ∞, n0 < n, nσ(n0) < nσ(n), stutf(nσ) ⊦ ∃ m. nσ(m) = nσ(n0) + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 76 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 77:

  ⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ(m) → (∃ m0. m0 ≤ m ∧ nσ(m0) = n)

• Proof step 78:

  # nσ = ∞, n0 < n, nσ(n0) < nσ(n), stutf(nσ), stutf(nσ) ∧ ⌜n⌝ < # nσ ∧ nσ(n0) + 1 ≤ nσ(n) → (∃ m0. m0 ≤ n ∧ nσ(m0) = nσ(n0) + 1) ⊦ ∃ m. nσ(m) = nσ(n0) + 1

Rule argument

Additional information

Used terms: nσ, n, nσ(n0) + 1

Proof step 77 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma stutf-alltaken

Proof step 78 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 79:

  # nσ = ∞, nσ(n1) = nσ(n0) + 1, n0 < n, n1 ≤ n, nσ(n0) < nσ(n), nσ(n1) ≤ nσ(n), stutf(nσ) ⊦ ∃ m. nσ(m) = nσ(n1)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat ⊦ ⌜m⌝ < ∞; (s) ⊦ m + 1 ≤ n ↔ m < n; (s)

Proof step 79 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used terms: n1

Proof step 80 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 81:

  leninv(nσ) ≠ ∞, nσ(((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to leninv(nσ).n)(n) + m + 1) ≠ n + 1, ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to leninv(nσ).n)(n) + m + 1 < ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to leninv(nσ).n)(n + 1), ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ⌜x⌝ (n) = x(n); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic

Proof step 81 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 82:

  leninv(nσ) = ⌜n0⌝ , leninv(nσ) ≠ ∞, nσ(((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to n0)(n) + m + 1) ≠ n + 1, ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to n0)(n) + m + 1 < ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to n0)(n + 1), ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: leninv(nσ), n0

Proof step 82 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 83:

  leninv(nσ) = ⌜n0⌝ , nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n + 1).n, n + 1 < n0, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s) m < n ⊦ (nf to n)(m) = nf(m); (s) ⊦ m ≤ n → (n < m ↔ false); (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m ≤ n → (⌜n⌝ + n∞0 < ⌜m⌝ ↔ false); (s) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 83 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 84:

  leninv(nσ) = ⌜n0⌝ , n0 = n1.+1, nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n + 1).n, n + 1 < n0, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: Nat freely generated by 0, +1; n0 < n ⊦ ¬ m + (* n) < (* n0); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic

Proof step 84 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 85:

  leninv(nσ) = ⌜n1⌝ + ⌜1⌝ , nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n + 1).n, n < n1, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m.+1 = m + 1; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 85 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 86:

  ⊦ stutf(nσ) → (leninv(nσ) = ⌜n⌝ + ⌜1⌝ ↔ (# nσ = ∞ ⊃ ∃ m. nσ(m) = n ∧ (∀ m0. m < m0 → nσ(m0) = n + 1);nσ((# nσ).n - 1) = n))

• Proof step 87:

  nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n + 1).n, n < n1, stutf(nσ), (# nσ = ∞ ⊃ ∃ m. nσ(m) = n1 ∧ (∀ m0. m < m0 → nσ(m0) = n1 + 1);nσ((# nσ).n - 1) = n1) ⊦

Rule argument

Additional information

Used terms: n1, nσ

Proof step 86 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma leninv-succ

Proof step 87 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 88:

  nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n + 1).n, n < n1, stutf(nσ), (# nσ = ∞ ⊃ nσ(m0) = n1 ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = n1) ⊦

Rule argument

Additional information

Proof step 88 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 89:

  # nσ = ∞, nσ(m0) = n1, nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n + 1).n, n < n1, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

• Proof step 147:

  nσ((# nσ).n - 1) = n1, # nσ ≠ ∞, nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n + 1).n, n < n1, stutf(nσ) ⊦

Rule argument

Additional information

Proof step 89 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 90:

  # nσ = ∞, nσ(max∞(λ n0. nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. nσ(n0) = n).n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, n < nσ(m0), stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ⌜m⌝ < ∞; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic

Proof step 90 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 91:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ∞, nσ(max∞(λ n0. nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. nσ(n0) = n).n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, n < nσ(m0), stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

• Proof step 106:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) ≠ ∞, nσ(max∞(λ n0. nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. nσ(n0) = n).n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, n < nσ(m0), stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used formulas: max∞(λ n0. nσ(n0) = n) = ∞

Proof step 91 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 92:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ∞, nσ(∞.n + m + 1) ≠ n + 1, n < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Proof step 92 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 93:

  ⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))

• Proof step 94:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ n + 1, n < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, (∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = n) ∨ (∀ n0. nσ(n0) ≠ n) ⊦

Rule argument

Additional information

Used terms: λ n0. nσ(n0) = n

Proof step 93 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma maxinf-inf

Proof step 94 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 95:

  ⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ(m) → (∃ m0. m0 ≤ m ∧ nσ(m0) = n)

• Proof step 96:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ n + 1, n < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, (∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = n) ∨ (∀ n0. nσ(n0) ≠ n), stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ n ≤ nσ(m0) → (∃ m1. m1 ≤ m0 ∧ nσ(m1) = n) ⊦

Rule argument

Additional information

Used terms: nσ, m0, n

Proof step 95 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma stutf-alltaken

Proof step 96 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 97:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ nσ(m1) + 1, m1 ≤ m0, nσ(m1) < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m1) + 1).n, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, (∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = nσ(m1)) ∨ (∀ n0. nσ(n0) ≠ nσ(m1)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic ⊦ ⌜m⌝ < ∞; (s) ⊦ m < n → m ≤ n; (s)

Proof step 97 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 98:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ nσ(m1) + 1, m1 ≤ m0, nσ(m1) < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m1) + 1).n, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = nσ(m1) ⊦

• Proof step 105:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ nσ(m1) + 1, m1 ≤ m0, nσ(m1) < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m1) + 1).n, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ n0. nσ(n0) ≠ nσ(m1) ⊦

Rule argument

Additional information

Proof step 98 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 99:

  # nσ = ∞, nσ(∞.n + m + 1) ≠ nσ(m1) + 1, m1 ≤ m0, nσ(m1) < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m1) + 1).n, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∃ n1. m0 < n1 ∧ nσ(n1) = nσ(m1) ⊦

Rule argument

Additional information

Used terms: m0

Proof step 99 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 100:

  # nσ = ∞, nσ(n1) = nσ(m1), nσ(∞.n + m + 1) ≠ nσ(m1) + 1, m0 < n1, m1 < n1, m1 ≤ m0, nσ(m1) < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m1) + 1).n, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat

Proof step 100 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 101:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

• Proof step 102:

  # nσ = ∞, nσ(n1) = nσ(m1), nσ(∞.n + m + 1) ≠ nσ(m1) + 1, m0 < n1, m1 ≤ m0, m1 < n1, nσ(m1) < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m1) + 1).n, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, mon≤(nσ) ∧ m0 ≤ m1 ∧ ⌜m1⌝ < # nσ → nσ(m0) ≤ nσ(m1) ⊦

Rule argument

Additional information

Used terms: nσ, m0, m1

Proof step 101 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 102 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 103:

  # nσ = ∞, nσ(n1) = nσ(m1), nσ(∞.n + m + 1) ≠ nσ(m1) + 1, m0 < n1, m1 < n1, m1 ≤ m0, nσ(m1) < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ(n0) = nσ(m1) + 1).n, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → m ≤ n; (s) ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)

Proof step 103 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 104:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → m ≤ n; (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: nσ, m0, n1

Proof step 104 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 105 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m1

Proof step 106 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 107:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , max∞(λ n0. nσ(n0) = n) ≠ ∞, nσ(n1 + m + 1) ≠ n + 1, n < nσ(m0), n1 + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. nσ(n0) = n), n1

Proof step 107 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 108:

  # nσ = ∞, max∞(λ n0. nσ(n0) = n) = ⌜n1⌝ , nσ(n1 + m + 1) ≠ n + 1, n < nσ(m0), n1 + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 108 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 109:

  ⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)

• Proof step 110:

  # nσ = ∞, nσ(n1) = n, nσ(n1 + m + 1) ≠ n + 1, n < nσ(m0), n1 + m + 1 < max∞(λ n0. nσ(n0) = n + 1).n, stutf(nσ), ∀ m. nσ(m) = n → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used terms: n1, λ n0. nσ(n0) = n

Proof step 109 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma maxinf-fin

Proof step 110 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 111:

  # nσ = ∞, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < max∞(λ n0. nσ(n0) = nσ(n1) + 1).n, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 111 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 112:

  # nσ = ∞, max∞(λ n0. nσ(n0) = nσ(n1) + 1) = ∞, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < max∞(λ n0. nσ(n0) = nσ(n1) + 1).n, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

• Proof step 123:

  # nσ = ∞, max∞(λ n0. nσ(n0) = nσ(n1) + 1) ≠ ∞, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < max∞(λ n0. nσ(n0) = nσ(n1) + 1).n, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used formulas: max∞(λ n0. nσ(n0) = nσ(n1) + 1) = ∞

Proof step 112 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 113:

  # nσ = ∞, max∞(λ n0. nσ(n0) = nσ(n1) + 1) = ∞, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < ∞.n, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 113 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 114:

  ⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))

• Proof step 115:

  # nσ = ∞, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < ∞.n, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, (∀ n. ∃ n0. n < n0 ∧ nσ(n0) = nσ(n1) + 1) ∨ (∀ n. nσ(n) ≠ nσ(n1) + 1) ⊦

Rule argument

Additional information

Used terms: λ n0. nσ(n0) = nσ(n1) + 1

Proof step 114 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma maxinf-inf

Proof step 115 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 116:

  ⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ(m) → (∃ m0. m0 ≤ m ∧ nσ(m0) = n)

• Proof step 117:

  # nσ = ∞, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < ∞.n, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, (∀ n. ∃ n0. n < n0 ∧ nσ(n0) = nσ(n1) + 1) ∨ (∀ n. nσ(n) ≠ nσ(n1) + 1), stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ nσ(n1) + 1 ≤ nσ(m0) → (∃ m1. m1 ≤ m0 ∧ nσ(m1) = nσ(n1) + 1) ⊦

Rule argument

Additional information

Used terms: nσ, m0, nσ(n1) + 1

Proof step 116 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma stutf-alltaken

Proof step 117 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 118:

  # nσ = ∞, nσ(m1) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(m1), m1 ≤ m0, n1 + m + 1 < ∞.n, nσ(n1) < nσ(m0), nσ(m1) ≤ nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, (∀ n. ∃ n0. n < n0 ∧ nσ(n0) = nσ(m1)) ∨ (∀ n. nσ(n) ≠ nσ(m1)) ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat ⊦ ⌜m⌝ < ∞; (s) ⊦ m + 1 ≤ n ↔ m < n; (s)

Proof step 118 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 119:

  # nσ = ∞, nσ(m1) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(m1), m1 ≤ m0, n1 + m + 1 < ∞.n, nσ(m1) ≤ nσ(m0), nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ n. ∃ n0. n < n0 ∧ nσ(n0) = nσ(m1) ⊦

• Proof step 122:

  # nσ = ∞, nσ(m1) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(m1), m1 ≤ m0, n1 + m + 1 < ∞.n, nσ(m1) ≤ nσ(m0), nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ n. nσ(n) ≠ nσ(m1) ⊦

Rule argument

Additional information

Proof step 119 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 120:

  # nσ = ∞, nσ(m1) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(m1), m1 ≤ m0, n1 + m + 1 < ∞.n, nσ(n1) < nσ(m0), nσ(m1) ≤ nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∃ n0. m0 < n0 ∧ nσ(n0) = nσ(m1) ⊦

Rule argument

Additional information

Used terms: m0

Proof step 120 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 121:

  # nσ = ∞, nσ(m1) = nσ(n1) + 1, nσ(n0) = nσ(m1), nσ(n1 + m + 1) ≠ nσ(m1), m0 < n0, m1 < n0, m1 ≤ m0, n1 + m + 1 < ∞.n, nσ(n1) < nσ(m0), nσ(m1) ≤ nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat

Proof step 121 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m = n + 1 → ¬ m ≤ n; (s) Used terms: n0

Proof step 122 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m1

Proof step 123 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 124:

  # nσ = ∞, max∞(λ n0. nσ(n0) = nσ(n1) + 1) = ⌜n⌝ , max∞(λ n0. nσ(n0) = nσ(n1) + 1) ≠ ∞, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < n, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. nσ(n0) = nσ(n1) + 1), n

Proof step 124 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 125:

  # nσ = ∞, max∞(λ n0. nσ(n0) = nσ(n1) + 1) = ⌜n⌝ , nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < n, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 125 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 126:

  ⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)

• Proof step 127:

  # nσ = ∞, nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < n, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. nσ(m) = nσ(n1) + 1 → ¬ n < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used terms: n, λ n0. nσ(n0) = nσ(n1) + 1

Proof step 126 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma maxinf-fin

Proof step 127 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 128:

  # nσ = ∞, nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n1 + m + 1 < n, nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n) → ¬ n < m, ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat

Proof step 128 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 129:

  # nσ = ∞, nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n1 + m + 1 < n, nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. nσ(m) = nσ(n) → ¬ n < m, m0 < n → nσ(n) = nσ(m0) + 1, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used terms: n

Proof step 129 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 130:

  # nσ = ∞, nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), nσ(n) ≠ nσ(m0) + 1, n ≤ m0, n1 + m + 1 < n, n1 + m + 1 < m0, nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. nσ(m) = nσ(n) → ¬ n < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m = n + 1 → ¬ m ≤ n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic

Proof step 130 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 131:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

• Proof step 132:

  # nσ = ∞, nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), nσ(n) ≠ nσ(m0) + 1, n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n) → ¬ n < m, ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, mon≤(nσ) ∧ m0 ≤ n1 ∧ ⌜n1⌝ < # nσ → nσ(m0) ≤ nσ(n1) ⊦

Rule argument

Additional information

Used terms: nσ, m0, n1

Proof step 131 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 132 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 133:

  # nσ = ∞, nσ(n) = nσ(n1) + 1, nσ(n) ≠ nσ(m0) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, n1 < m0, nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n) → ¬ n < m, ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m ≤ n ↔ ¬ n < m; (ss) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ¬ m + n < m; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat ⊦ m < n → (n ≤ m ↔ false); (s)

Proof step 133 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 134:

  # nσ = ∞, nσ(n) = nσ(n1) + 1, nσ(n) ≠ nσ(m0) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 < m0, n1 + m + 1 < n, n1 + m + 1 < m0, nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. nσ(m) = nσ(n) → ¬ n < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, nσ(n1 + 1) = nσ(n1) → ¬ n1 < n1 + 1 ⊦

Rule argument

Additional information

Used terms: n1 + 1

Proof step 134 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 135:

  # nσ = ∞, nσ(n) = nσ(n1) + 1, nσ(n) ≠ nσ(m0) + 1, nσ(n1 + m + 1) ≠ nσ(n), nσ(n1 + 1) ≠ nσ(n1), n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, n1 < m0, nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n) → ¬ n < m, ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s)

Proof step 135 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 136:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

• Proof step 137:

  # nσ = ∞, nσ(n) = nσ(n1) + 1, nσ(n) ≠ nσ(m0) + 1, nσ(n1 + m + 1) ≠ nσ(n), nσ(n1 + 1) ≠ nσ(n1), n ≤ m0, n1 < m0, n1 + m + 1 < n, n1 + m + 1 < m0, nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. nσ(m) = nσ(n) → ¬ n < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, mon≤(nσ) ∧ n1 + 1 ≤ n1 + m + 1 ∧ ⌜n1 + m + 1⌝ < # nσ → nσ(n1 + 1) ≤ nσ(n1 + m + 1) ⊦

Rule argument

Additional information

Used terms: nσ, n1 + 1, n1 + m + 1

Proof step 136 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 137 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 138:

  # nσ = ∞, nσ(n) = nσ(n1) + 1, nσ(n1 + 1) ≠ nσ(n1), nσ(n1 + m + 1) ≠ nσ(n), nσ(n) ≠ nσ(m0) + 1, n ≤ m0, n1 < m0, n1 + m + 1 < n, n1 + m + 1 < m0, nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), nσ(n1 + 1) ≤ nσ(n1 + m + 1), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. nσ(m) = nσ(n) → ¬ n < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ m ≤ m + n; (s) ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)

Proof step 138 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 139:

  # nσ = ∞, nσ(n) = nσ(n1) + 1, nσ(n1 + 1) = nσ(n1) + 1, nσ(n1 + 1) ≠ nσ(n1), nσ(n1 + m + 1) ≠ nσ(n), nσ(n) ≠ nσ(m0) + 1, n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, n1 < m0, nσ(n1 + 1) ≤ nσ(n1 + m + 1), nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n) → ¬ n < m, ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

• Proof step 142:

  # nσ = ∞, nσ(n) = nσ(n1) + 1, nσ(n1 + 1) ≠ nσ(n1) + 1, nσ(n1 + 1) ≠ nσ(n1), nσ(n1 + m + 1) ≠ nσ(n), nσ(n) ≠ nσ(m0) + 1, n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, n1 < m0, nσ(n1 + 1) ≤ nσ(n1 + m + 1), nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n) → ¬ n < m, ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used formulas: nσ(n1 + 1) = nσ(n1) + 1

Proof step 139 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 140:

  # nσ = ∞, nσ(n1 + 1) = nσ(n), nσ(n) = nσ(n1) + 1, nσ(n) ≠ nσ(n1), nσ(n1 + m + 1) ≠ nσ(n), nσ(n) ≠ nσ(m0) + 1, n ≤ m0, n1 < m0, n1 + m + 1 < n, n1 + m + 1 < m0, nσ(n) < nσ(n1 + m + 1), nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. nσ(m) = nσ(n) → ¬ n < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic

Proof step 140 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 141:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) ⊦ m < n → ¬ n < m; (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ ¬ m < n → ¬ m + m0 < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: nσ, n1 + m + 1, n

Proof step 141 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 142 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 143:

  # nσ = ∞, nσ(n) = nσ(n1) + 1, nσ(n1 + 1) ≠ nσ(n), nσ(n1 + 1) ≠ nσ(n1), nσ(n1 + m + 1) ≠ nσ(n), nσ(n) ≠ nσ(m0) + 1, n ≤ m0, n1 < m0, n1 + m + 1 < n, n1 + m + 1 < m0, nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), nσ(n1 + 1) ≤ nσ(n1 + m + 1), stutf(nσ), ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. nσ(m) = nσ(n) → ¬ n < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1 ⊦

Rule argument

Additional information

Proof step 143 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 144:

  ⊦ stutf(nσ) ↔ nσ ≠ [] ∧ nσ(0) = 0 ∧ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1)

• Proof step 145:

  # nσ = ∞, nσ(0) = 0, nσ(n) = nσ(n1) + 1, nσ(n1 + 1) ≠ nσ(n), nσ(n1 + 1) ≠ nσ(n1), nσ(n1 + m + 1) ≠ nσ(n), nσ(n) ≠ nσ(m0) + 1, nσ ≠ [], n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, n1 < m0, nσ(n1 + 1) ≤ nσ(n1 + m + 1), nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), ∀ m. nσ(m) = nσ(n) → ¬ n < m, ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1 ⊦

Rule argument

Additional information

Used terms: nσ

Proof step 144 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma stutf-def

Proof step 145 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 146:

  # nσ = ∞, nσ(0) = 0, nσ(n) = nσ(n1) + 1, nσ ≠ [], nσ(n) ≠ nσ(m0) + 1, nσ(n1 + m + 1) ≠ nσ(n), nσ(n1 + 1) ≠ nσ(n1), nσ(n1 + 1) ≠ nσ(n), n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, n1 < m0, nσ(n1 + 1) ≤ nσ(n1 + m + 1), nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), ∀ m. nσ(m) = nσ(n) → ¬ n < m, ∀ m. nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ n0. nσ(n0 + 1) = nσ(n0) ∨ nσ(n0 + 1) = nσ(n0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s)

Proof step 146 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n1

Proof step 147 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 148:

  # nσ ≠ ∞, nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n + 1).n, n < nσ((# nσ).n - 1), stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 148 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 149:

  # nσ = ⌜n1⌝ , # nσ ≠ ∞, nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n + m + 1 < max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n + 1).n, n < nσ(n1 - 1), stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: # nσ, n1

Proof step 149 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 150:

  # nσ = ⌜n1⌝ , nσ(max∞(λ n0. n0 < n1 ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. n0 < n1 ∧ nσ(n0) = n).n + m + 1 < max∞(λ n0. n0 < n1 ∧ nσ(n0) = n + 1).n, n < nσ(n1 - 1), stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 150 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 151:

  # nσ = ⌜n1⌝ , n1 = m0 + 1, n1 ≠ 0, nσ(max∞(λ n0. n0 < n1 ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. n0 < n1 ∧ nσ(n0) = n).n + m + 1 < max∞(λ n0. n0 < n1 ∧ nσ(n0) = n + 1).n, n < nσ(m0), stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: stutf-nonempty: ⊦ stutf(nσ) → (nσ = [] ↔ false); ⊦ # σ = ⌜0⌝ ↔ σ = []; (s) n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n1, m0

Proof step 151 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 152:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n).n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n + 1).n, n < nσ(m0), stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m + 1 ≠ 0; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 152 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 153:

  # nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n) = ∞, nσ(max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n).n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n + 1).n, n < nσ(m0), stutf(nσ) ⊦

• Proof step 163:

  # nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n) ≠ ∞, nσ(max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n).n + m + 1) ≠ n + 1, max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n).n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n + 1).n, n < nσ(m0), stutf(nσ) ⊦

Rule argument

Additional information

Used formulas: max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n) = ∞

Proof step 153 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 154:

  # nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n) = ∞, nσ(∞.n + m + 1) ≠ n + 1, n < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n + 1).n ⊦

Rule argument

Additional information

Proof step 154 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 155:

  ⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))

• Proof step 156:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(∞.n + m + 1) ≠ n + 1, n < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n + 1).n, (∀ n0. ∃ n1. n0 < n1 ∧ ¬ m0 < n1 ∧ nσ(n1) = n) ∨ (∀ n0. ¬(¬ m0 < n0 ∧ nσ(n0) = n)) ⊦

Rule argument

Additional information

Used terms: λ n0. ¬ m0 < n0 ∧ nσ(n0) = n

Proof step 155 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma maxinf-inf

Proof step 156 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 157:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(∞.n + m + 1) ≠ n + 1, n < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n + 1).n, (∀ n0. ∃ n1. n0 < n1 ∧ ¬ m0 < n1 ∧ nσ(n1) = n) ∨ (∀ n0. m0 < n0 ∨ nσ(n0) ≠ n) ⊦

Rule argument

Additional information

Proof step 157 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 158:

  ⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ(m) → (∃ m0. m0 ≤ m ∧ nσ(m0) = n)

• Proof step 159:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(∞.n + m + 1) ≠ n + 1, n < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n + 1).n, (∀ n0. ∃ n1. n0 < n1 ∧ ¬ m0 < n1 ∧ nσ(n1) = n) ∨ (∀ n0. m0 < n0 ∨ nσ(n0) ≠ n), stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ n ≤ nσ(m0) → (∃ m1. m1 ≤ m0 ∧ nσ(m1) = n) ⊦

Rule argument

Additional information

Used terms: nσ, m0, n

Proof step 158 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma stutf-alltaken

Proof step 159 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 160:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(∞.n + m + 1) ≠ nσ(m1) + 1, m1 ≤ m0, nσ(m1) < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(m1) + 1).n, (∀ n0. ∃ n1. n0 < n1 ∧ ¬ m0 < n1 ∧ nσ(n1) = nσ(m1)) ∨ (∀ n0. m0 < n0 ∨ nσ(n0) ≠ nσ(m1)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic ⊦ m < n → m ≤ n; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf

Proof step 160 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 161:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(∞.n + m + 1) ≠ nσ(m1) + 1, m1 ≤ m0, nσ(m1) < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(m1) + 1).n, ∀ n0. ∃ n1. n0 < n1 ∧ ¬ m0 < n1 ∧ nσ(n1) = nσ(m1) ⊦

• Proof step 162:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(∞.n + m + 1) ≠ nσ(m1) + 1, m1 ≤ m0, nσ(m1) < nσ(m0), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(m1) + 1).n, ∀ n0. m0 < n0 ∨ nσ(n0) ≠ nσ(m1) ⊦

Rule argument

Additional information

Proof step 161 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m0

Proof step 162 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: m1

Proof step 163 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 164:

  # nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n) = ⌜n1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n) ≠ ∞, nσ(n1 + m + 1) ≠ n + 1, n < nσ(m0), n1 + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n + 1).n, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n), n1

Proof step 164 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 165:

  # nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n) = ⌜n1⌝ , nσ(n1 + m + 1) ≠ n + 1, n < nσ(m0), n1 + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n + 1).n, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 165 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 166:

  ⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)

• Proof step 167:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n1) = n, nσ(n1 + m + 1) ≠ n + 1, n < nσ(m0), n1 + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = n + 1).n, stutf(nσ), ¬ m0 < n1, ∀ m. ¬ m0 < m ∧ nσ(m) = n → ¬ n1 < m ⊦

Rule argument

Additional information

Used terms: n1, λ n0. ¬ m0 < n0 ∧ nσ(n0) = n

Proof step 166 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma maxinf-fin

Proof step 167 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 168:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 ≤ m0, n1 + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1).n, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 168 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 169:

  # nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1) = ∞, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1).n, n1 ≤ m0, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m ⊦

• Proof step 179:

  # nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1) ≠ ∞, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1).n, n1 ≤ m0, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m ⊦

Rule argument

Additional information

Used formulas: max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1) = ∞

Proof step 169 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 170:

  # nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1) = ∞, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < ∞.n, n1 ≤ m0, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 170 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 171:

  ⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))

• Proof step 172:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 ≤ m0, n1 + m + 1 < ∞.n, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, (∀ n. ∃ n0. n < n0 ∧ ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1) ∨ (∀ n. ¬(¬ m0 < n ∧ nσ(n) = nσ(n1) + 1)) ⊦

Rule argument

Additional information

Used terms: λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1

Proof step 171 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma maxinf-inf

Proof step 172 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 173:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < ∞.n, n1 ≤ m0, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, (∀ n. ∃ n0. n < n0 ∧ ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1) ∨ (∀ n. m0 < n ∨ nσ(n) ≠ nσ(n1) + 1) ⊦

Rule argument

Additional information

Proof step 173 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 174:

  ⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ(m) → (∃ m0. m0 ≤ m ∧ nσ(m0) = n)

• Proof step 175:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 ≤ m0, n1 + m + 1 < ∞.n, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, (∀ n. ∃ n0. n < n0 ∧ ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1) ∨ (∀ n. m0 < n ∨ nσ(n) ≠ nσ(n1) + 1), stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ nσ(n1) + 1 ≤ nσ(m0) → (∃ m1. m1 ≤ m0 ∧ nσ(m1) = nσ(n1) + 1) ⊦

Rule argument

Additional information

Used terms: nσ, m0, nσ(n1) + 1

Proof step 174 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma stutf-alltaken

Proof step 175 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 176:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(m1) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(m1), m1 ≤ m0, n1 + m + 1 < ∞.n, n1 ≤ m0, nσ(n1) < nσ(m0), nσ(m1) ≤ nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, (∀ n. ∃ n0. n < n0 ∧ ¬ m0 < n0 ∧ nσ(n0) = nσ(m1)) ∨ (∀ n. m0 < n ∨ nσ(n) ≠ nσ(m1)) ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf

Proof step 176 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 177:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(m1) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(m1), m1 ≤ m0, n1 ≤ m0, n1 + m + 1 < ∞.n, nσ(m1) ≤ nσ(m0), nσ(n1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, ∀ n. ∃ n0. n < n0 ∧ ¬ m0 < n0 ∧ nσ(n0) = nσ(m1) ⊦

• Proof step 178:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(m1) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(m1), m1 ≤ m0, n1 ≤ m0, n1 + m + 1 < ∞.n, nσ(m1) ≤ nσ(m0), nσ(n1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, ∀ n. m0 < n ∨ nσ(n) ≠ nσ(m1) ⊦

Rule argument

Additional information

Proof step 177 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m0

Proof step 178 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: m1

Proof step 179 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 180:

  # nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1) = ⌜n⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1) ≠ ∞, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 ≤ m0, n1 + m + 1 < n, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1), n

Proof step 180 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 181:

  # nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1) = ⌜n⌝ , nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 ≤ m0, n1 + m + 1 < n, nσ(n1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 181 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 182:

  ⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)

• Proof step 183:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n1) + 1, n1 + m + 1 < n, n1 ≤ m0, nσ(n1) < nσ(m0), stutf(nσ), ¬ m0 < n, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) + 1 → ¬ n < m ⊦

Rule argument

Additional information

Used terms: n, λ n0. ¬ m0 < n0 ∧ nσ(n0) = nσ(n1) + 1

Proof step 182 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma maxinf-fin

Proof step 183 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 184:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 ≤ m0, n1 + m + 1 < n, n1 + m + 1 < m0, nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 184 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 185:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

• Proof step 186:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, n1 ≤ m0, nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m, mon≤(nσ) ∧ m0 ≤ n1 ∧ ⌜n1⌝ < # nσ → nσ(m0) ≤ nσ(n1) ⊦

Rule argument

Additional information

Used terms: nσ, m0, n1

Proof step 185 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 186 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 187:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 ≤ m0, n1 + m + 1 < n, n1 + m + 1 < m0, nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → m ≤ n; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m ≤ n → (n < m ↔ false); (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)

Proof step 187 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 188:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

• Proof step 189:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, n1 ≤ m0, nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m, mon≤(nσ) ∧ n1 + m + 1 ≤ n ∧ ⌜n⌝ < # nσ → nσ(n1 + m + 1) ≤ nσ(n) ⊦

Rule argument

Additional information

Used terms: nσ, n1 + m + 1, n

Proof step 188 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 189 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 190:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m < m0, n1 + m < n, n1 ≤ m0, n1 + m + 1 < n, n1 + m + 1 < m0, nσ(n1 + m + 1) < nσ(m0), nσ(n1 + m + 1) < nσ(n), nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), nσ(n1 + m + 1) ≤ nσ(n1), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n ↔ ¬ n ≤ m; (ss) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m < n → n ≠ 0; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ¬ m < n → ¬ m + m0 < n; (s) ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) ⊦ m ≤ n → (n < m ↔ false); (s) ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ m + 1 ≤ n ↔ m < n; (s)

Proof step 190 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 191:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

• Proof step 192:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, n1 ≤ m0, n1 + m < n, n1 + m < m0, nσ(n1 + m + 1) ≤ nσ(n1), nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), nσ(n1 + m + 1) < nσ(n), nσ(n1 + m + 1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m, mon≤(nσ) ∧ m0 ≤ n1 + m + 1 ∧ ⌜n1 + m + 1⌝ < # nσ → nσ(m0) ≤ nσ(n1 + m + 1) ⊦

Rule argument

Additional information

Used terms: nσ, m0, n1 + m + 1

Proof step 191 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 192 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 193:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m < m0, n1 + m < n, n1 ≤ m0, n1 + m + 1 < n, n1 + m + 1 < m0, nσ(n1 + m + 1) < nσ(m0), nσ(n1 + m + 1) < nσ(n), nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), nσ(n1 + m + 1) ≤ nσ(n1), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → m ≤ n; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)

Proof step 193 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 194:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

• Proof step 195:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, n1 ≤ m0, n1 + m < n, n1 + m < m0, nσ(n1 + m + 1) ≤ nσ(n1), nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), nσ(n1 + m + 1) < nσ(n), nσ(n1 + m + 1) < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m, mon≤(nσ) ∧ n1 + 1 ≤ n1 + m + 1 ∧ ⌜n1 + m + 1⌝ < # nσ → nσ(n1 + 1) ≤ nσ(n1 + m + 1) ⊦

Rule argument

Additional information

Used terms: nσ, n1 + 1, n1 + m + 1

Proof step 194 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 195 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 196:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m < m0, n1 + m < n, n1 ≤ m0, n1 + m + 1 < n, n1 + m + 1 < m0, nσ(n1 + m + 1) < nσ(m0), nσ(n1 + m + 1) < nσ(n), nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), nσ(n1 + m + 1) ≤ nσ(n1), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, ⌜n1⌝ + ⌜m⌝ + ⌜1⌝ < # nσ → nσ(n1 + 1) ≤ nσ(n1 + m + 1) ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ m ≤ m + n; (s) ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)

Proof step 196 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 197:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, n1 ≤ m0, n1 + m < n, n1 + m < m0, nσ(n1 + m + 1) ≤ nσ(n1), nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), nσ(n1 + m + 1) < nσ(n), nσ(n1 + m + 1) < nσ(m0), stutf(nσ), ¬ ⌜n1⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m ⊦

• Proof step 198:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, n1 ≤ m0, n1 + m < n, n1 + m < m0, nσ(n1 + m + 1) ≤ nσ(n1), nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), nσ(n1 + m + 1) < nσ(n), nσ(n1 + m + 1) < nσ(m0), nσ(n1 + 1) ≤ nσ(n1 + m + 1), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m ⊦

Rule argument

Additional information

Proof step 197 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule subst equation

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ + n∞1 < ⌜n⌝ + n∞2 ↔ n∞1 < n∞2; (s) ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → (n ≤ m ↔ false); (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf

Proof step 198 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 199:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m < m0, n1 + m < n, n1 ≤ m0, n1 + m + 1 < n, n1 + m + 1 < m0, nσ(n1 + 1) ≤ nσ(n1 + m + 1), nσ(n1 + m + 1) < nσ(m0), nσ(n1 + m + 1) < nσ(n), nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), nσ(n1 + m + 1) ≤ nσ(n1), nσ(n1 + 1) ≤ nσ(n1), nσ(n1 + 1) < nσ(m0), nσ(n1 + 1) < nσ(n), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m ≤ n → (n < m ↔ false); (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat

Proof step 199 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 200:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

• Proof step 201:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, n1 ≤ m0, n1 + m < n, n1 + m < m0, nσ(n1 + 1) < nσ(n), nσ(n1 + 1) < nσ(m0), nσ(n1 + 1) ≤ nσ(n1), nσ(n1 + m + 1) ≤ nσ(n1), nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), nσ(n1 + m + 1) < nσ(n), nσ(n1 + m + 1) < nσ(m0), nσ(n1 + 1) ≤ nσ(n1 + m + 1), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m, mon≤(nσ) ∧ m0 ≤ n1 + 1 ∧ ⌜n1 + 1⌝ < # nσ → nσ(m0) ≤ nσ(n1 + 1) ⊦

Rule argument

Additional information

Used terms: nσ, m0, n1 + 1

Proof step 200 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq

Proof step 201 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 202:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m < m0, n1 + m < n, n1 ≤ m0, n1 + m + 1 < n, n1 + m + 1 < m0, nσ(n1 + 1) ≤ nσ(n1 + m + 1), nσ(n1 + m + 1) < nσ(m0), nσ(n1 + m + 1) < nσ(n), nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), nσ(n1 + m + 1) ≤ nσ(n1), nσ(n1 + 1) ≤ nσ(n1), nσ(n1 + 1) < nσ(m0), nσ(n1 + 1) < nσ(n), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → m ≤ n; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ n∞0 + ⌜1⌝ = n∞ → (⌜n⌝ + ⌜1⌝ < n∞ ↔ ⌜n⌝ < n∞0); (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 202 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 203:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m + 1 < m0, n1 + m + 1 < n, n1 ≤ m0, n1 + m < n, n1 + m < m0, nσ(n1 + 1) < nσ(n), nσ(n1 + 1) < nσ(m0), nσ(n1 + 1) ≤ nσ(n1), nσ(n1 + m + 1) ≤ nσ(n1), nσ(n) ≤ nσ(m0), nσ(n1) < nσ(m0), nσ(n1 + m + 1) < nσ(n), nσ(n1 + m + 1) < nσ(m0), nσ(n1 + 1) ≤ nσ(n1 + m + 1), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m, ¬ m0 < n1 + 1 ∧ nσ(n1 + 1) = nσ(n1) → ¬ n1 < n1 + 1 ⊦

Rule argument

Additional information

Used terms: n1 + 1

Proof step 203 from proof inv-get-inverse-between (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 204:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(n) = nσ(n1) + 1, nσ(n1 + 1) ≠ nσ(n1), nσ(n1 + m + 1) ≠ nσ(n), n ≤ m0, n1 + m < m0, n1 + m < n, n1 + m + 1 < n, n1 + m + 1 < m0, n1 < m0, nσ(n1 + 1) < nσ(n1), nσ(n1 + 1) < nσ(m0), nσ(n1 + 1) < nσ(n), nσ(n1 + 1) ≤ nσ(n1 + m + 1), nσ(n1 + m + 1) < nσ(m0), nσ(n1 + m + 1) < nσ(n), nσ(n1) < nσ(m0), nσ(n) ≤ nσ(m0), nσ(n1 + m + 1) ≤ nσ(n1), nσ(n1 + 1) + 1 < nσ(m0), stutf(nσ), ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n) → ¬ n < m, ∀ m. ¬ m0 < m ∧ nσ(m) = nσ(n1) → ¬ n1 < m ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m.+1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) lf: ⊦ m < n ∧ n < n0 → m.+1 < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) ⊦ m < n → m ≤ n; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ¬ m < n → ¬ m + m0 < n; (s) ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s)

Proof step 204 from proof inv-get-inverse-between (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 205:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(n)

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m ≤ m + n; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ n∞0 + ⌜1⌝ = n∞ → (⌜n⌝ + ⌜1⌝ < n∞ ↔ ⌜n⌝ < n∞0); (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) ⊦ m < n → ¬ n < m; (s) Used terms: nσ, n1, n1 + 1

Proof step 205 from proof inv-get-inverse-between (previous step) (back to tree)

Lemma monle-leq