Proof step 1 from proof leninv-inf (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

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

Rule argument

Additional information

Proof step 2 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

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

Rule argument

Additional information

Proof step 3 from proof leninv-inf (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ leninv(nσ) = max∞(λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ ¬(# nσ = ∞ ∧ (∀ m0. m < m0 → nσ(m0) = n - 1))))

• Proof step 5:

  stutf(nσ) ⊦ max∞(λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ ¬(# nσ = ∞ ∧ (∀ m0. m < m0 → nσ(m0) = n - 1)))) = ∞ ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0))

Rule argument

Additional information

Used terms: nσ

Proof step 4 from proof leninv-inf (previous step) (back to tree)

Lemma leninv-def

Proof step 5 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  stutf(nσ) ⊦ max∞(λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n - 1)))) = ∞ ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0))

Rule argument

Additional information

Proof step 6 from proof leninv-inf (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

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

• Proof step 8:

  stutf(nσ) ⊦ (∀ n. ∃ n0. n < n0 ∧ (n0 ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n0 - 1))))) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n - 1))))) ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0))

Rule argument

Additional information

Used terms: λ n. n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n - 1)))

Proof step 7 from proof leninv-inf (previous step) (back to tree)

Lemma maxinf-inf

Proof step 8 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  stutf(nσ) ⊦ (∀ n. ∃ n0. n < n0 ∧ (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n0 - 1)))) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n - 1))))) ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0))

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)

Proof step 9 from proof leninv-inf (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 10:

  # nσ = ∞, stutf(nσ) ⊦ (∀ n. ∃ n0. n < n0 ∧ (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n0 - 1)))) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n - 1))))) ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0))

• Proof step 50:

  # nσ ≠ ∞, stutf(nσ) ⊦ (∀ n. ∃ n0. n < n0 ∧ (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n0 - 1)))) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ(m0) = n - 1))))) ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0))

Rule argument

Additional information

Used formulas: # nσ = ∞

Proof step 10 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  # nσ = ∞, stutf(nσ) ⊦ (∀ n. ∃ n0. n < n0 ∧ (∃ m. nσ(m) = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n0 - 1))) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. nσ(m) = n - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n - 1)))) ↔ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0))

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

Proof step 11 from proof leninv-inf (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 12:

  # nσ = ∞, stutf(nσ), (∀ n. ∃ n0. n < n0 ∧ (∃ m. nσ(m) = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n0 - 1))) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. nσ(m) = n - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n - 1)))) ⊦ ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)

• Proof step 23:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0) ⊦ ∀ n. ∃ n0. n < n0 ∧ (∃ m. nσ(m) = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n0 - 1)), ∀ n. ¬(n ≠ 0 → (∃ m. nσ(m) = n - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n - 1)))

Rule argument

Additional information

Proof step 12 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  # nσ = ∞, stutf(nσ), (∀ n. ∃ n0. n < n0 ∧ (∃ m. nσ(m) = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n0 - 1))) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. nσ(m) = n - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n - 1)))) ⊦ ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)

Rule argument

Additional information

Proof step 13 from proof leninv-inf (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 14:

  # nσ = ∞, stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ m. nσ(m) = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n0 - 1)) ⊦ ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)

• Proof step 22:

  # nσ = ∞, stutf(nσ), ∀ n. ¬(n ≠ 0 → (∃ m. nσ(m) = n - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n - 1))) ⊦ ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)

Rule argument

Additional information

Proof step 14 from proof leninv-inf (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 15:

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

Rule argument

Additional information

Used terms: nσ(m) + 1

Proof step 15 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  # nσ = ∞, nσ(m0) = n0 - 1, nσ(m1) ≠ n0 - 1, m0 < m1, nσ(m) + 1 < n0, stutf(nσ) ⊦ ∃ 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

Proof step 16 from proof leninv-inf (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 17:

  # nσ = ∞, n0 = m2 + 1, nσ(m0) = m2, n0 ≠ 0, nσ(m1) ≠ m2, m0 < m1, nσ(m) + 1 < n0, stutf(nσ) ⊦ ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)

Rule argument

Additional information

Used simplifier rules: 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 n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n0, m2

Proof step 17 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  # nσ = ∞, nσ(m1) ≠ nσ(m0), m0 < m1, nσ(m) < nσ(m0), stutf(nσ) ⊦ ∃ 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 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 18 from proof leninv-inf (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 19:

  # nσ = ∞, nσ(m1) ≠ nσ(m0), m0 < m1, nσ(m) < nσ(m0), stutf(nσ) ⊦ m < m0 ∧ nσ(m) < nσ(m0)

Rule argument

Additional information

Used terms: m0

Proof step 19 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  # nσ = ∞, nσ(m1) ≠ nσ(m0), m0 < m1, m0 ≤ m, nσ(m) < nσ(m0), stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 20 from proof leninv-inf (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 21:

  ⊦ 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 → (n ≤ m ↔ false); (s) ⊦ ⌜m⌝ < ∞; (s) Used terms: nσ, m0, m

Proof step 21 from proof leninv-inf (previous step) (back to tree)

Lemma monle-leq

Proof step 22 from proof leninv-inf (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: 0

Proof step 23 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), n0 ≠ 0 → nσ(m) = n0 - 1 ∧ ¬(m < m0 → nσ(m0) = n0 - 1) ⊦ ∃ n0. n < n0 ∧ (∃ m. nσ(m) = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n0 - 1))

Rule argument

Additional information

Proof step 24 from proof leninv-inf (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 25:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n. ∃ m. nσ(m) = n, n0 ≠ 0 → nσ(m) = n0 - 1 ∧ ¬(m < m0 → nσ(m0) = n0 - 1) ⊦ ∃ n0. n < n0 ∧ (∃ m. nσ(m) = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n0 - 1))

• Proof step 37:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), n0 ≠ 0 → nσ(m) = n0 - 1 ∧ ¬(m < m0 → nσ(m0) = n0 - 1) ⊦ ∀ n. ∃ m. nσ(m) = n, ∃ n0. n < n0 ∧ (∃ m. nσ(m) = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n0 - 1))

Rule argument

Additional information

Used formulas: ∀ n. ∃ m. nσ(m) = n

Proof step 25 from proof leninv-inf (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 26:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n. ∃ m. nσ(m) = n, n0 ≠ 0 → nσ(m) = n0 - 1 ∧ ¬(m < m0 → nσ(m0) = n0 - 1) ⊦ n < n + 1 ∧ (∃ m. nσ(m) = n + 1 - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n + 1 - 1))

Rule argument

Additional information

Used terms: n + 1

Proof step 26 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n. ∃ m. nσ(m) = n, n0 ≠ 0 → nσ(m) = n0 - 1 ∧ ¬(m < m0 → nσ(m0) = n0 - 1) ⊦ ∃ m1. nσ(m1) = n ∧ ¬(∀ m2. m1 < m2 → nσ(m2) = 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 ⊦ n + 0 = n; (s) ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s) ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s)

Proof step 27 from proof leninv-inf (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 28:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∃ m. nσ(m) = n, ∀ n. ∃ m. nσ(m) = n, n0 ≠ 0 → nσ(m) = n0 - 1 ∧ ¬(m < m0 → nσ(m0) = n0 - 1) ⊦ ∃ m1. nσ(m1) = n ∧ ¬(∀ m2. m1 < m2 → nσ(m2) = nσ(m1))

Rule argument

Additional information

Used terms: n

Proof step 28 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n. ∃ m. nσ(m) = n, n0 ≠ 0 → nσ(m) = n0 - 1 ∧ ¬(m < m0 → nσ(m0) = n0 - 1) ⊦ ∃ m. nσ(m) = nσ(m1) ∧ ¬(∀ m2. m < m2 → nσ(m2) = nσ(m))

Rule argument

Additional information

Proof step 29 from proof leninv-inf (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 30:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n. ∃ m. nσ(m) = n, n0 ≠ 0 → nσ(m) = n0 - 1 ∧ ¬(m < m0 → nσ(m0) = n0 - 1) ⊦ nσ(m1) = nσ(m1) ∧ ¬(∀ m2. m1 < m2 → nσ(m2) = nσ(m1))

Rule argument

Additional information

Used terms: m1

Proof step 30 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m2. m1 < m2 → nσ(m2) = nσ(m1), ∀ n. ∃ m. nσ(m) = n, n0 ≠ 0 → nσ(m) = n0 - 1 ∧ ¬(m < m0 → nσ(m0) = n0 - 1) ⊦

Rule argument

Additional information

Proof step 31 from proof leninv-inf (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 32:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∃ m. nσ(m) = nσ(m1) + 1, ∀ m2. m1 < m2 → nσ(m2) = nσ(m1), ∀ n. ∃ m. nσ(m) = n, n0 ≠ 0 → nσ(m) = n0 - 1 ∧ ¬(m < m0 → nσ(m0) = n0 - 1) ⊦

Rule argument

Additional information

Used terms: nσ(m1) + 1

Proof step 32 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  # nσ = ∞, nσ(m2) = nσ(m1) + 1, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ m2. m1 < m2 → nσ(m2) = nσ(m1), ∀ n. ∃ m. nσ(m) = n, n0 ≠ 0 → nσ(m) = n0 - 1 ∧ ¬(m < m0 → nσ(m0) = n0 - 1) ⊦

Rule argument

Additional information

Proof step 33 from proof leninv-inf (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 34:

  # nσ = ∞, nσ(m2) = nσ(m1) + 1, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), m1 < m2 → nσ(m2) = nσ(m1), ∀ n. ∃ m. nσ(m) = n, n0 ≠ 0 → nσ(m) = n0 - 1 ∧ ¬(m < m0 → nσ(m0) = n0 - 1) ⊦

Rule argument

Additional information

Used terms: m2

Proof step 34 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 35:

  # nσ = ∞, nσ(m2) = nσ(m1) + 1, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), m1 < m2 → nσ(m2) = nσ(m1), ∀ n. ∃ m. nσ(m) = n, n0 ≠ 0 → nσ(m) = n0 - 1 ∧ ¬(m < m0 → nσ(m0) = n0 - 1) ⊦

Rule argument

Additional information

Proof step 35 from proof leninv-inf (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 36:

  ⊦ 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 + 1 → ¬ m ≤ n; (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m = m + n ↔ n = 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: nσ, m2, m1

Proof step 36 from proof leninv-inf (previous step) (back to tree)

Lemma monle-leq

Proof step 37 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 38:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), n0 ≠ 0 → nσ(m) = n0 - 1 ∧ ¬(m < m0 → nσ(m0) = n0 - 1) ⊦ ∃ m. nσ(m) = n1, ∃ n0. n < n0 ∧ (∃ m. nσ(m) = n0 - 1 ∧ ¬(∀ m0. m < m0 → nσ(m0) = n0 - 1))

Rule argument

Additional information

Proof step 38 from proof leninv-inf (previous step) (back to tree)

User applied rule weakening

And obtained the following new goals

• Proof step 39:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0) ⊦ ∃ m. nσ(m) = n1

Rule argument

Additional information

Proof step 39 from proof leninv-inf (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 40:

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

• Proof step 41:

  # 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 40 from proof leninv-inf (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 41 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 42:

  # 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 42 from proof leninv-inf (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 43:

  # 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 43 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

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

Rule argument

Additional information

Proof step 44 from proof leninv-inf (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 45:

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

Rule argument

Additional information

Used terms: m

Proof step 45 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 46:

  # nσ = ∞, m < m0, nσ(m) < nσ(m0), stutf(nσ) ⊦ ∃ m0. nσ(m0) = nσ(m) + 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 46 from proof leninv-inf (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 47:

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

• Proof step 48:

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

Rule argument

Additional information

Used terms: nσ, m0, nσ(m) + 1

Proof step 47 from proof leninv-inf (previous step) (back to tree)

Lemma stutf-alltaken

Proof step 48 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 49:

  # nσ = ∞, nσ(m1) = nσ(m) + 1, m < m0, m1 ≤ m0, nσ(m) < nσ(m0), nσ(m1) ≤ nσ(m0), stutf(nσ) ⊦ ∃ m0. nσ(m0) = 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 49 from proof leninv-inf (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used terms: m1

Proof step 50 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 51:

  # nσ ≠ ∞, stutf(nσ), (∀ n. ∃ n0. n < n0 ∧ (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n0 - 1)) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1))) ⊦

Rule argument

Additional information

Proof step 51 from proof leninv-inf (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 52:

  ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

• Proof step 53:

  stutf(nσ), (∀ n. ∃ n0. n < n0 ∧ (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n0 - 1)) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1))), ∃ n. # nσ = ⌜n⌝ ⊦

Rule argument

Additional information

Used terms: # nσ

Proof step 52 from proof leninv-inf (previous step) (back to tree)

A spec-lemma

Proof step 53 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 54:

  # nσ = ⌜n⌝ , stutf(nσ), (∀ n. ∃ n0. n < n0 ∧ (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n0 - 1)) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1))) ⊦

Rule argument

Additional information

Proof step 54 from proof leninv-inf (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 55:

  # nσ = ⌜n⌝ , stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n0 - 1) ⊦

• Proof step 63:

  # nσ = ⌜n⌝ , stutf(nσ), ∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1)) ⊦

Rule argument

Additional information

Proof step 55 from proof leninv-inf (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 56:

  # nσ = ⌜n⌝ , n = n1.+1, stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n0 - 1) ⊦

Rule argument

Additional information

Used simplifier rules: stutf-nonempty: ⊦ stutf(nσ) → (nσ = [] ↔ false); Nat freely generated by 0, +1; ⊦ # σ = ⌜0⌝ ↔ σ = []; (s)

Proof step 56 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 57:

  # nσ = ⌜n1⌝ + ⌜1⌝ , stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ m. ¬ n1 < m ∧ nσ(m) = n0 - 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 ⊦ 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⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ m.+1 = m + 1; (s)

Proof step 57 from proof leninv-inf (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 58:

  # nσ = ⌜n1⌝ + ⌜1⌝ , stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ m. ¬ n1 < m ∧ nσ(m) = n0 - 1), ∃ n0. nσ(n1) + 1 < n0 ∧ (∃ m. ¬ n1 < m ∧ nσ(m) = n0 - 1) ⊦

Rule argument

Additional information

Used terms: nσ(n1) + 1

Proof step 58 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 59:

  # nσ = ⌜n1⌝ + ⌜1⌝ , nσ(m) = n0 - 1, m ≤ n1, nσ(n1) + 1 < n0, stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ m. ¬ n1 < m ∧ nσ(m) = n0 - 1) ⊦

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 59 from proof leninv-inf (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 60:

  # nσ = ⌜n1⌝ + ⌜1⌝ , n0 = m0 + 1, nσ(m) = m0, n0 ≠ 0, m ≤ n1, nσ(n1) + 1 < n0, stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ m. ¬ n1 < m ∧ nσ(m) = n0 - 1) ⊦

Rule argument

Additional information

Used simplifier rules: 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 n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n0, m0

Proof step 60 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 61:

  # nσ = ⌜n1⌝ + ⌜1⌝ , m ≤ n1, nσ(n1) < nσ(m), stutf(nσ), ∀ n. ∃ n0. n < n0 ∧ (∃ m. ¬ n1 < m ∧ nσ(m) = n0 - 1) ⊦

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 61 from proof leninv-inf (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 62:

  ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ(m) ≤ nσ(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∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ¬ n < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: nσ, m, n1

Proof step 62 from proof leninv-inf (previous step) (back to tree)

Lemma monle-leq

Proof step 63 from proof leninv-inf (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 64:

  # nσ = ⌜n⌝ , n = n0.+1, stutf(nσ), ∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ(m) = n - 1)) ⊦

Rule argument

Additional information

Used simplifier rules: stutf-nonempty: ⊦ stutf(nσ) → (nσ = [] ↔ false); Nat freely generated by 0, +1; ⊦ # σ = ⌜0⌝ ↔ σ = []; (s)

Proof step 64 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 65:

  # nσ = ⌜n0⌝ + ⌜1⌝ , stutf(nσ), ∀ n. ¬(n ≠ 0 → (∃ m. ¬ n0 < m ∧ nσ(m) = n - 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 ⊦ 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⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ m.+1 = m + 1; (s)

Proof step 65 from proof leninv-inf (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 66:

  # nσ = ⌜n0⌝ + ⌜1⌝ , nσ(n0) + 1 ≠ 0, stutf(nσ), ∀ n. ¬(n ≠ 0 → (∃ m. ¬ n0 < m ∧ nσ(m) = n - 1)) ⊦ ∃ m. ¬ n0 < m ∧ nσ(m) = nσ(n0) + 1 - 1

Rule argument

Additional information

Used terms: nσ(n0) + 1

Proof step 66 from proof leninv-inf (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 67:

  # nσ = ⌜n0⌝ + ⌜1⌝ , stutf(nσ), ∀ n. ¬(n ≠ 0 → (∃ m. ¬ n0 < m ∧ nσ(m) = n - 1)) ⊦ ∃ m. ¬ n0 < m ∧ nσ(m) = nσ(n0)

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 + 0 = n; (s) ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s) ⊦ m + 1 ≠ 0; (s)

Proof step 67 from proof leninv-inf (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s) Used terms: n0