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, nσ[m] + 1 < n0, m0 < m1, 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 = ∞, nσ[m0] = m2, n0 = m2 + 1, nσ[m1] ≠ m2, n0 ≠ 0, 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], nσ[m] < nσ[m0], m0 < m1, 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], nσ[m] < nσ[m0], m0 < m1, stutf(nσ), m0 ≤ m ⊦

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

• Proof step 37:

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

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

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

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

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

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), ∀ 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 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 = ∞, nσ[m] < nσ[m0], m < 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, nσ[m] < nσ[m0], stutf(nσ), m1 ≤ m0, nσ[m1] ≤ nσ[m0] ⊦ ∃ 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. # nσ = ⌜n⌝ , (∀ 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

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-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) ⊦ 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, nσ[n1] + 1 < n0, stutf(nσ), m ≤ n1, ∀ 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⌝ , nσ[m] = m0, n0 = m0 + 1, n0 ≠ 0, nσ[n1] + 1 < n0, stutf(nσ), m ≤ n1, ∀ 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⌝ , nσ[n1] < nσ[m], stutf(nσ), m ≤ n1, ∀ 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-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 ⊦ 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-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) ⊦ 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