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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

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

Rule argument

Additional information

Proof step 3 from proof inv-get-inverse (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)) ≠ n, ⌜n⌝ < leninv(nσ), stutf(nσ) ⊦

Rule argument

Additional information

Used terms: nσ

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

Lemma inv-def

Proof step 5 from proof inv-get-inverse (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)) ≠ n, ⌜n⌝ < leninv(nσ), stutf(nσ) ⊦

Rule argument

Additional information

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

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

User applied rule case distinction

And obtained the following new goals

• Proof step 7:

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

• Proof step 43:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

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

Rule argument

Additional information

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

Proof step 8 from proof inv-get-inverse (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 = ∞, nσ(max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n) ≠ n, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0) ⊦

Rule argument

Additional information

Used terms: nσ

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

Lemma leninv-inf

Proof step 10 from proof inv-get-inverse (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) ≠ n, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

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

User applied rule cut formula

And obtained the following new goals

• Proof step 12:

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

• Proof step 39:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

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

Rule argument

Additional information

Proof step 13 from proof inv-get-inverse (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:

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

Rule argument

Additional information

Used terms: λ n0. nσ(n0) = n

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

Lemma maxinf-inf

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

User applied rule case distinction

And obtained the following new goals

• Proof step 16:

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

• Proof step 27:

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

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 17:

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

Rule argument

Additional information

Used terms: 0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  # nσ = ∞, n1 ≠ 0, nσ(∞.n) ≠ nσ(n1), stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n0. ∃ n2. n0 < n2 ∧ nσ(n2) = nσ(n1) ⊦

Rule argument

Additional information

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

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

User applied rule all left

And obtained the following new goals

• Proof step 19:

  # nσ = ∞, n1 ≠ 0, nσ(∞.n) ≠ nσ(n1), stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∃ m0. n1 < m0 ∧ nσ(n1) < nσ(m0), ∀ n0. ∃ n2. n0 < n2 ∧ nσ(n2) = nσ(n1) ⊦

Rule argument

Additional information

Used terms: n1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  # nσ = ∞, n1 ≠ 0, nσ(∞.n) ≠ nσ(n1), n1 < n, nσ(n1) < nσ(n), stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n0. ∃ n2. n0 < n2 ∧ nσ(n2) = nσ(n1) ⊦

Rule argument

Additional information

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

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

User applied rule all left

And obtained the following new goals

• Proof step 21:

  # nσ = ∞, n1 ≠ 0, nσ(∞.n) ≠ nσ(n1), n1 < n, nσ(n1) < nσ(n), stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∃ n2. n < n2 ∧ nσ(n2) = nσ(n1) ⊦

Rule argument

Additional information

Used terms: n

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  # nσ = ∞, nσ(n2) = nσ(n1), n1 ≠ 0, nσ(∞.n) ≠ nσ(n1), n < n2, n1 < n2, n1 < n, n1 + 1 < n2, nσ(n1) < nσ(n), stutf(nσ), ∀ 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 22 from proof inv-get-inverse (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 23:

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

• Proof step 24:

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

Rule argument

Additional information

Used terms: nσ, n, n1

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

Lemma monle-leq

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  # nσ = ∞, nσ(n2) = nσ(n1), n1 ≠ 0, nσ(∞.n) ≠ nσ(n1), n < n2, n1 < n2, n1 < n, n1 + 1 < n2, nσ(n1) < nσ(n), stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(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 25 from proof inv-get-inverse (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 26:

  ⊦ 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, n, n2

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

Lemma monle-leq

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

User applied rule weakening

And obtained the following new goals

• Proof step 28:

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

Rule argument

Additional information

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

User applied rule structural induction

And obtained the following new goals

• Proof step 29:

  # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0), ∀ n0. nσ(n0) ≠ 0 ⊦

• Proof step 30:

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

Rule argument

Additional information

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

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

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

Proof step 30 from proof inv-get-inverse (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), ∀ n1. nσ(n1) ≠ n0 + 1, ∀ nσ0. # nσ0 ≠ ∞ ∨ ¬ stutf(nσ0) ∨ ¬(∀ m. ∃ m0. m < m0 ∧ nσ0(m) < nσ0(m0)) ∨ ¬(∀ n1. nσ0(n1) ≠ n0) ⊦

Rule argument

Additional information

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

Proof step 31 from proof inv-get-inverse (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), ∀ n1. nσ(n1) ≠ n0 + 1, # nσ ≠ ∞ ∨ ¬ stutf(nσ) ∨ ¬(∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)) ∨ ¬(∀ n1. nσ(n1) ≠ n0) ⊦

Rule argument

Additional information

Used terms: nσ

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

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

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 34:

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

Rule argument

Additional information

Used terms: n1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 35:

  # nσ = ∞, n1 < n, nσ(n1) < nσ(n), stutf(nσ), ∀ n. nσ(n) ≠ nσ(n1) + 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 35 from proof inv-get-inverse (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 36:

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

• Proof step 37:

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

Rule argument

Additional information

Used terms: nσ, n, nσ(n1) + 1

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

Lemma stutf-alltaken

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 38:

  # nσ = ∞, nσ(n0) = nσ(n1) + 1, n0 ≤ n, n1 < n, nσ(n1) < nσ(n), nσ(n0) ≤ nσ(n), stutf(nσ), ∀ n. nσ(n) ≠ 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 eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat ⊦ ⌜m⌝ < ∞; (s) ⊦ m + 1 ≤ n ↔ m < n; (s)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n0

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 40:

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

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 40 from proof inv-get-inverse (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) = n) = ⌜n1⌝ , nσ(n1) ≠ n, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 42:

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

Rule argument

Additional information

Used terms: n1, λ n0. nσ(n0) = n

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

Lemma maxinf-fin

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

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

Rule argument

Additional information

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

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 45:

  leninv(nσ) = ⌜n0⌝ , leninv(nσ) ≠ ∞, nσ(((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n) to n0)(n)) ≠ n, ⌜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 45 from proof inv-get-inverse (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 46:

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

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

User applied rule cut formula

And obtained the following new goals

• Proof step 47:

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

• Proof step 80:

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

Rule argument

Additional information

Used formulas: max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 48:

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

Rule argument

Additional information

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

User applied rule weakening

And obtained the following new goals

• Proof step 49:

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

Rule argument

Additional information

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

User applied rule constructor cut

And obtained the following new goals

• Proof step 50:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 51:

  leninv(nσ) = ⌜n1⌝ + ⌜1⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, 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 51 from proof inv-get-inverse (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 52:

  ⊦ 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 53:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = 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 52 from proof inv-get-inverse (previous step) (back to tree)

Lemma leninv-succ

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 54:

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

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 55:

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

• Proof step 66:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 56:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 57:

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

• Proof step 58:

  # nσ = ∞, n ≤ nσ(n0), stutf(nσ), ∀ m0. n0 < m0 → nσ(m0) = nσ(n0) + 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 57 from proof inv-get-inverse (previous step) (back to tree)

Lemma maxinf-inf

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

User applied rule case distinction

And obtained the following new goals

• Proof step 59:

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

• Proof step 62:

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

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 60:

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

Rule argument

Additional information

Used terms: n0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 61:

  # nσ = ∞, n0 < n1, nσ(n1) ≤ nσ(n0), stutf(nσ), ∀ m0. n0 < m0 → nσ(m0) = 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 61 from proof inv-get-inverse (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

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 → ¬ m ≤ n; (s) Used terms: n1

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 63:

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

• Proof step 64:

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

Rule argument

Additional information

Used terms: nσ, n0, n

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

Lemma stutf-alltaken

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 65:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 67:

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

Rule argument

Additional information

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 68:

  # nσ = ⌜n1⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, # 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 68 from proof inv-get-inverse (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 69:

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

Rule argument

Additional information

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

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 70:

  # nσ = ⌜n1⌝ , max∞(λ n0. n0 < n1 ∧ nσ(n0) = n) = ∞, n1 = n2 + 1, n1 ≠ 0, n ≤ nσ(n2), 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, n2

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 71:

  # nσ = ⌜n2⌝ + ⌜1⌝ , max∞(λ n0. ¬ n2 < n0 ∧ nσ(n0) = n) = ∞, n ≤ nσ(n2), stutf(nσ) ⊦

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 + 1 ≠ 0; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 72:

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

• Proof step 73:

  # nσ = ⌜n2⌝ + ⌜1⌝ , n ≤ nσ(n2), stutf(nσ), (∀ n0. ∃ n1. n0 < n1 ∧ ¬ n2 < n1 ∧ nσ(n1) = n) ∨ (∀ n0. ¬(¬ n2 < n0 ∧ nσ(n0) = n)) ⊦

Rule argument

Additional information

Used terms: λ n0. ¬ n2 < n0 ∧ nσ(n0) = n

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

Lemma maxinf-inf

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 74:

  # nσ = ⌜n2⌝ + ⌜1⌝ , n ≤ nσ(n2), stutf(nσ), (∀ n0. ∃ n1. n0 < n1 ∧ ¬ n2 < n1 ∧ nσ(n1) = n) ∨ (∀ n0. n2 < n0 ∨ nσ(n0) ≠ n) ⊦

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 75:

  # nσ = ⌜n2⌝ + ⌜1⌝ , n ≤ nσ(n2), stutf(nσ), ∀ n0. ∃ n1. n0 < n1 ∧ ¬ n2 < n1 ∧ nσ(n1) = n ⊦

• Proof step 76:

  # nσ = ⌜n2⌝ + ⌜1⌝ , n ≤ nσ(n2), stutf(nσ), ∀ n0. n2 < n0 ∨ nσ(n0) ≠ n ⊦

Rule argument

Additional information

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n2

Proof step 76 from proof inv-get-inverse (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 = ⌜n2⌝ + ⌜1⌝ , n ≤ nσ(n2), stutf(nσ), ∀ n0. n2 < n0 ∨ nσ(n0) ≠ n, stutf(nσ) ∧ ⌜n2⌝ < # nσ ∧ n ≤ nσ(n2) → (∃ m0. m0 ≤ n2 ∧ nσ(m0) = n) ⊦

Rule argument

Additional information

Used terms: nσ, n2, n

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

Lemma stutf-alltaken

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 79:

  # nσ = ⌜n2⌝ + ⌜1⌝ , n0 ≤ n2, nσ(n0) ≤ nσ(n2), stutf(nσ), ∀ n1. n2 < n1 ∨ nσ(n1) ≠ 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 < 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 79 from proof inv-get-inverse (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: n0

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 81:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 82:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

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

User applied rule constructor cut

And obtained the following new goals

• Proof step 83:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 84:

  leninv(nσ) = ⌜n2⌝ + ⌜1⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ⌜n1⌝ , nσ(n1) ≠ n, n ≤ n2, 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 84 from proof inv-get-inverse (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 85:

  ⊦ 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 86:

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

Rule argument

Additional information

Used terms: n2, nσ

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

Lemma leninv-succ

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 87:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ⌜n1⌝ , nσ(n1) ≠ n, n ≤ n2, stutf(nσ), (# nσ = ∞ ⊃ nσ(n0) = n2 ∧ (∀ m0. n0 < m0 → nσ(m0) = nσ(n0) + 1);nσ((# nσ).n - 1) = n2) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 88:

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

Rule argument

Additional information

Used terms: n1, λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n

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

Lemma maxinf-fin