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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  nσ(n + 1) = nσ(n) + 1, inv(nσ)(nσ(n)) ≠ n, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  nσ(n + 1) = nσ(n) + 1, inv(nσ)(nσ(n)) ≠ n, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦

Rule argument

Additional information

Proof step 3 from proof inv-inverts (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:

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

Rule argument

Additional information

Used terms: nσ

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

Lemma inv-def

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

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

Rule argument

Additional information

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

User applied rule cut formula

And obtained the following new goals

• Proof step 7:

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

• Proof step 39:

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

Rule argument

Additional information

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

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

User applied rule case distinction

And obtained the following new goals

• Proof step 8:

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

Rule argument

Additional information

Used simplifier rules: leninv-inf-strong: ⊦ stutf(nσ) → ( leninv(nσ) = ∞ ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)) ∧ (∀ m. ∃ m0. m < nσ(m0))); ⊦ ⌜nf⌝ (n) = nf(n); (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ ⌜n⌝ .n = n; (s)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

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

Rule argument

Additional information

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

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 10:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

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

User applied rule constructor cut

And obtained the following new goals

• Proof step 12:

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

• Proof step 15:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  leninv(nσ) = ⌜0⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)) = ⌜n⌝ , nσ(n + 1) = nσ(n) + 1, [](nσ(n)) ≠ n, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ nf.take(0) = []; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

  ⊦ stutf(nσ) → (leninv(nσ) = ⌜0⌝ ↔ nσ = ⌜λ n. 0⌝ )

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ (n) = nf(n); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: nσ

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

Lemma leninv-zero

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

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

Rule argument

Additional information

Used simplifier rules: ⊦ m.+1 = m + 1; (s) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 17:

  ⊦ 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 18:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = nσ(n)) = ⌜n⌝ , nσ(n + 1) = nσ(n) + 1, (λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n).take(n2 + 1)(nσ(n)) ≠ n, ⌜n⌝ + ⌜1⌝ < # nσ, 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 17 from proof inv-inverts (previous step) (back to tree)

Lemma leninv-succ

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

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

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 20:

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

• Proof step 32:

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

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

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

Rule argument

Additional information

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

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

User applied rule all left

And obtained the following new goals

• Proof step 22:

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

Rule argument

Additional information

Used terms: n

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

User applied rule cut formula

And obtained the following new goals

• Proof step 23:

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

• Proof step 25:

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

Rule argument

Additional information

Used formulas: m < n

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  # nσ = ∞, max∞(λ n0. nσ(n0) = nσ(n)) = ⌜n⌝ , nσ(n) = nσ(m) + 1, nσ(n + 1) = nσ(n) + 1, (λ m0. max∞(λ n0. nσ(n0) = m0).n).take(nσ(n))(nσ(n)) ≠ n, m < n, stutf(nσ), ∀ m0. m < m0 → nσ(m0) = 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 24 from proof inv-inverts (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m < n → ¬ n < m; (s) eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (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: n + 1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

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

Rule argument

Additional information

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

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

User applied rule all left

And obtained the following new goals

• Proof step 27:

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

Rule argument

Additional information

Used terms: n + 1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

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

Rule argument

Additional information

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

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

User applied rule cut formula

And obtained the following new goals

• Proof step 29:

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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m ≤ n → (n ≤ m ↔ m = n); (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 ⊦ nf.take(n)(m) = nf(m); (s) ⊦ ⌜n⌝ .n = n; (s) Used formulas: n < m

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

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

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 31:

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

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

Lemma monle-leq

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

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

Rule argument

Additional information

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 34:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 35:

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

Rule argument

Additional information

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

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 36:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

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

Rule argument

Additional information

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

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 38:

  ⊦ 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) 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) ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m ≤ n → (n < m ↔ false); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic m < n ⊦ nf.take(n)(m) = nf(m); (s) ⊦ ⌜n⌝ .n = n; (s) Used terms: nσ, n, m0

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

Lemma monle-leq

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

User applied rule cut formula

And obtained the following new goals

• Proof step 40:

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

• Proof step 50:

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

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ ≠ ∞; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 42:

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

• Proof step 43:

  nσ(n + 1) = nσ(n) + 1, (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n).take(leninv(nσ).n)⌝ ) (nσ(n)) ≠ n, ⌜n⌝ + ⌜1⌝ < # nσ, stutf(nσ), (∀ 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 42 from proof inv-inverts (previous step) (back to tree)

Lemma maxinf-inf

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

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

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 45:

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

• Proof step 49:

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

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 46:

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

Rule argument

Additional information

Used terms: n + 1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 47:

  nσ(n + 1) = nσ(n) + 1, nσ(n1) = nσ(n), (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⌝ + ⌜1⌝ < # nσ, n + 1 < n1, ⌜n1⌝ < # 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 47 from proof inv-inverts (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 48:

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

Rule argument

Additional information

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

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

Lemma monle-leq

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2; (forward) from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic 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 Used terms: n

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 51:

  ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

• Proof step 52:

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

Rule argument

Additional information

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

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 53:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜n⌝ = ⌜n0⌝ ↔ n = n0; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 54:

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

• Proof step 55:

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

Rule argument

Additional information

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

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

Lemma maxinf-fin

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 56:

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

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 57:

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

Rule argument

Additional information

Used terms: n0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 58:

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

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 59:

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

• Proof step 60:

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

Rule argument

Additional information

Used terms: nσ, n + 1, n0

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

Lemma monle-leq

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 61:

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

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≠ n → (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 ⊦ ¬ n < n; (s) ⊦ m + 1 ≤ n ↔ m < n; (s)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2; (forward) from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic 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 Used terms: n