Proof step 1 from proof inv-inverts-leq (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(inv(nσ)(nσ(n)) < n ↔ false), ⌜n⌝ < # nσ, ⌜nσ(n)⌝ < leninv(nσ), stutf(nσ) ⊦

Rule argument

Additional information

Proof step 2 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  inv(nσ)(nσ(n)) < n, ⌜n⌝ < # nσ, ⌜nσ(n)⌝ < 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-inverts-leq (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⌝ .take(leninv(nσ).n))

• Proof step 5:

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

Rule argument

Additional information

Used terms: nσ

Proof step 4 from proof inv-inverts-leq (previous step) (back to tree)

Lemma inv-def

Proof step 5 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ .take(n) = ⌜nf.take(n)⌝ ; (s)

Proof step 6 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 7:

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

• Proof step 31:

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

Rule argument

Additional information

Proof step 7 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s) ⊦ ⌜nf⌝ (n) = nf(n); (s)

Proof step 8 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 9:

  ⊦ (max∞(np) = ∞ ⊃ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n));(∃ n. np(max∞(np).n)) ∧ (∀ m. np(m) → ¬ max∞(np) < ⌜m⌝ ))

• Proof step 10:

  leninv(nσ) = ∞, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)).n < n, ⌜n⌝ < # nσ, stutf(nσ), (max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)) = ∞ ⊃ (∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ(n1) = nσ(n)) ∨ (∀ n0. ¬(⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n))) ; (∃ n0. ⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)).n⌝ < # nσ ∧ nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)).n) = nσ(n)) ∧ (∀ m. ⌜m⌝ < # nσ ∧ nσ(m) = nσ(n) → ¬ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)) < ⌜m⌝ ) ) ⊦

Rule argument

Additional information

Used terms: λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)

Proof step 9 from proof inv-inverts-leq (previous step) (back to tree)

Lemma maxinf-def

Proof step 10 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  leninv(nσ) = ∞, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)).n < n, ⌜n⌝ < # nσ, stutf(nσ), (max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)) = ∞ ⊃ (∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ(n1) = nσ(n)) ∨ (∀ n0. ¬ ⌜n0⌝ < # nσ ∨ nσ(n0) ≠ nσ(n)) ; max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)) < # nσ ∧ nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)).n) = nσ(n) ∧ (∀ m. ⌜m⌝ < # nσ ∧ nσ(m) = nσ(n) → ¬ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)) < ⌜m⌝ ) ) ⊦

Rule argument

Additional information

Used simplifier rules: n∞ ≠ ∞ ⊦ ⌜n∞.n⌝ = n∞; (s)

Proof step 11 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 12:

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

• Proof step 27:

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

Rule argument

Additional information

Proof step 12 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

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

Rule argument

Additional information

Proof step 13 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

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

• Proof step 15:

  leninv(nσ) = ∞, ⌜n⌝ < # nσ, stutf(nσ), ∞.n < n, (∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ(n1) = nσ(n)) ∨ (∀ n0. ¬ ⌜n0⌝ < # nσ ∨ nσ(n0) ≠ nσ(n)), (∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ(n1) = nσ(n)) ∨ (∀ n0. ¬(⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n))) ⊦

Rule argument

Additional information

Used terms: λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)

Proof step 14 from proof inv-inverts-leq (previous step) (back to tree)

Lemma maxinf-inf

Proof step 15 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  leninv(nσ) = ∞, ⌜n⌝ < # nσ, stutf(nσ), ∞.n < n, (∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ(n1) = nσ(n)) ∨ (∀ n0. ¬ ⌜n0⌝ < # nσ ∨ nσ(n0) ≠ nσ(n)) ⊦

Rule argument

Additional information

Proof step 16 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 17:

  leninv(nσ) = ∞, ⌜n⌝ < # nσ, stutf(nσ), ∞.n < n, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ(n1) = nσ(n) ⊦

• Proof step 26:

  leninv(nσ) = ∞, ⌜n⌝ < # nσ, stutf(nσ), ∞.n < n, ∀ n0. ¬ ⌜n0⌝ < # nσ ∨ nσ(n0) ≠ nσ(n) ⊦

Rule argument

Additional information

Proof step 17 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 18:

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

• Proof step 19:

  # nσ = ∞, ⌜n⌝ < # nσ, stutf(nσ), ∞.n < n, ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ(n1) = nσ(n) ⊦

Rule argument

Additional information

Used terms: nσ

Proof step 18 from proof inv-inverts-leq (previous step) (back to tree)

Lemma leninv-inf

Proof step 19 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  # nσ = ∞, stutf(nσ), ∞.n < n, ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = nσ(n) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

Proof step 20 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 21:

  # nσ = ∞, stutf(nσ), ∞.n < n, ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∃ m0. n < m0 ∧ nσ(n) < nσ(m0), ∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = nσ(n) ⊦

Rule argument

Additional information

Used terms: n

Proof step 21 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

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

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 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 22 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 23:

  # nσ = ∞, n < m0, nσ(n) < nσ(m0), stutf(nσ), ∞.n + 1 < m0, ∞.n < n, ∞.n < m0, ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∃ n1. m0 < n1 ∧ nσ(n1) = nσ(n) ⊦

Rule argument

Additional information

Used terms: m0

Proof step 23 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  # nσ = ∞, nσ(n1) = nσ(n), m0 < n1, n < m0, n + 1 < n1, n < n1, nσ(n) < nσ(m0), stutf(nσ), ∞.n < n1, ∞.n + 1 < m0, ∞.n < n, ∞.n < m0, (∞.n + 1) + 1 < n1, ∞.n + 1 < n1, ∀ m. ∃ m0. m < m0 ∧ nσ(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 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 24 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 25:

  ⊦ 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 25 from proof inv-inverts-leq (previous step) (back to tree)

Lemma monle-leq

Proof step 26 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 27 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

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

Rule argument

Additional information

Proof step 28 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 29:

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

Rule argument

Additional information

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

Proof step 29 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 30 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 31 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 32:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ (n) = x(n); (s)

Proof step 32 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 33:

  leninv(nσ) = ⌜n0⌝ , leninv(nσ) ≠ ∞, (λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n).take(n0)(nσ(n)) < n, ⌜n⌝ < # nσ, ⌜nσ(n)⌝ < 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 33 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 34:

  leninv(nσ) = ⌜n0⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)).n < n, ⌜n⌝ < # nσ, nσ(n) < 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 ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) m < n ⊦ nf.take(n)(m) = nf(m); (s) ⊦ m ≤ n → (n < m ↔ false); (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 34 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 35:

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

Rule argument

Additional information

Used simplifier rules: Nat freely generated by 0, +1; ⊦ ¬ n < 0; (s)

Proof step 35 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 36:

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

Rule argument

Additional information

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

Proof step 36 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 37:

  ⊦ 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 38:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)).n < n, ⌜n⌝ < # nσ, nσ(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 37 from proof inv-inverts-leq (previous step) (back to tree)

Lemma leninv-succ

Proof step 38 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

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

Rule argument

Additional information

Proof step 39 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 40:

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

• Proof step 60:

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

Rule argument

Additional information

Proof step 40 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

Proof step 41 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 42:

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

• Proof step 55:

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

Rule argument

Additional information

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

Proof step 42 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 43:

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

Rule argument

Additional information

Proof step 43 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 44:

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

• Proof step 45:

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

Rule argument

Additional information

Used terms: λ n0. nσ(n0) = nσ(n)

Proof step 44 from proof inv-inverts-leq (previous step) (back to tree)

Lemma maxinf-inf

Proof step 45 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 46:

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

• Proof step 54:

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

Rule argument

Additional information

Proof step 46 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 47:

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

Rule argument

Additional information

Used terms: m + 1

Proof step 47 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 48:

  # nσ = ∞, nσ(n1) = nσ(n), m + 1 < n1, nσ(n) ≤ nσ(m), stutf(nσ), ∞.n < n, ∀ m0. m < m0 → nσ(m0) = nσ(m) + 1, ∀ n0. ∃ n1. n0 < n1 ∧ nσ(n1) = nσ(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 48 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 49:

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

Rule argument

Additional information

Used terms: m + 1

Proof step 49 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 50:

  # nσ = ∞, nσ(n1) = nσ(n), nσ(n0) = nσ(n), m + 1 < n1, m + 1 < n0, nσ(n) ≤ nσ(m), stutf(nσ), ∞.n < n, ∀ m0. m < 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 50 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 51:

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

• Proof step 52:

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

Rule argument

Additional information

Used terms: nσ, m + 1, n1

Proof step 51 from proof inv-inverts-leq (previous step) (back to tree)

Lemma monle-leq

Proof step 52 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 53:

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

Rule argument

Additional information

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

Proof step 53 from proof inv-inverts-leq (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) ⊦ ¬ n < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m.+1 = m + 1; (s) ⊦ m = n + 1 → ¬ m ≤ n; (s) Used terms: m.+1

Proof step 54 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 55 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 56:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 56 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 57:

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

• Proof step 58:

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

Rule argument

Additional information

Used terms: n1, λ n0. nσ(n0) = nσ(n)

Proof step 57 from proof inv-inverts-leq (previous step) (back to tree)

Lemma maxinf-fin

Proof step 58 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 59:

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

Rule argument

Additional information

Proof step 59 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 60 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 61:

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

Rule argument

Additional information

Proof step 61 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 62:

  # nσ = ⌜n1⌝ , # nσ ≠ ∞, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)).n < n, ⌜n⌝ < # nσ, nσ(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 62 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 63:

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

Rule argument

Additional information

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

Proof step 63 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 64:

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

Proof step 64 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 65:

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

Rule argument

Additional information

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

Proof step 65 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 66:

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

• Proof step 73:

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

Rule argument

Additional information

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

Proof step 66 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 67:

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

Rule argument

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat

Proof step 67 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 68:

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

• Proof step 69:

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

Rule argument

Additional information

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

Proof step 68 from proof inv-inverts-leq (previous step) (back to tree)

Lemma maxinf-inf

Proof step 69 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 70:

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

Rule argument

Additional information

Proof step 70 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 71:

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

• Proof step 72:

  # nσ = ⌜m⌝ + ⌜1⌝ , n ≤ m, nσ(n) ≤ nσ(m), stutf(nσ), ∞.n < m, ∞.n < n, ∀ n0. m < n0 ∨ nσ(n0) ≠ nσ(n) ⊦

Rule argument

Additional information

Proof step 71 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ 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: m + 1

Proof step 72 from proof inv-inverts-leq (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: n

Proof step 73 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 74:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat

Proof step 74 from proof inv-inverts-leq (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 75:

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

• Proof step 76:

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

Rule argument

Additional information

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

Proof step 75 from proof inv-inverts-leq (previous step) (back to tree)

Lemma maxinf-fin

Proof step 76 from proof inv-inverts-leq (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 77:

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

Rule argument

Additional information

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

Proof step 77 from proof inv-inverts-leq (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: n