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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ⌜n⌝ < # nσ, stutf(nσ) ⊦ ⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ⌜n⌝ < # nσ, stutf(nσ) ⊦ ⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))

Rule argument

Additional information

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

User applied rule cut formula

And obtained the following new goals

• Proof step 4:

  nσ(n) = 0, ⌜n⌝ < # nσ, stutf(nσ) ⊦ ⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))

• Proof step 16:

  nσ(n) ≠ 0, ⌜n⌝ < # nσ, stutf(nσ) ⊦ ⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))

Rule argument

Additional information

Used formulas: nσ(n) = 0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  nσ(n) = 0, ⌜n⌝ < # nσ, stutf(nσ) ⊦ ⌜0⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = 0)

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

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

• Proof step 7:

  nσ(n) = 0, ⌜n⌝ < # nσ, stutf(nσ) ⊦ nσ = ⌜λ n. 0⌝ ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = 0)

Rule argument

Additional information

Used terms: nσ

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

Lemma leninv-zero

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

User applied rule case distinction

And obtained the following new goals

• Proof step 8:

  # nσ = ∞, nσ(n) = 0, nσ ≠ ⌜λ n. 0⌝ , ⌜n⌝ < # nσ, stutf(nσ), ∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = 0 ⊦

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  # nσ = ∞, nσ(n) = 0, nσ ≠ ⌜λ n. 0⌝ , stutf(nσ), ∀ m. n < m → nσ(m) = 0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

  ⊦ σ0 = σ1 ↔ # σ0 = # σ1 ∧ (∀ n. ⌜n⌝ < # σ0 → σ0(n) = σ1(n))

• Proof step 11:

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

Rule argument

Additional information

Used terms: nσ, ⌜λ n. 0⌝ , ⌜λ n. 0⌝ , nσ

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  # nσ = ∞, nσ(n) = 0, nσ(n0) ≠ 0, stutf(nσ), ∀ m. n < m → nσ(m) = 0 ⊦

Rule argument

Additional information

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

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

User applied rule all left

And obtained the following new goals

• Proof step 13:

  # nσ = ∞, nσ(n) = 0, nσ(n0) ≠ 0, stutf(nσ), n < n0 → nσ(n0) = 0 ⊦

Rule argument

Additional information

Used terms: n0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  # nσ = ∞, nσ(n) = 0, nσ(n0) ≠ 0, n0 ≤ n, stutf(nσ) ⊦

Rule argument

Additional information

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

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 15:

  ⊦ 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⌝ < ∞; (s) ⊦ m ≤ 0 ↔ m = 0; (s) Used terms: nσ, n0, n

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

Lemma monle-leq

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

User applied rule structural induction

And obtained the following new goals

• Proof step 17:

  n1.+1 = nσ(n), nσ(n) ≠ 0, ⌜n⌝ < # nσ, stutf(nσ), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦ ⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  n1 + 1 = nσ(n), nσ(n) ≠ 0, ⌜n⌝ < # nσ, stutf(nσ), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦ ⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))

Rule argument

Additional information

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

Proof step 18 from proof eq-leninv (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 19:

  n1 + 1 = nσ(n), ⌜nσ(n)⌝ = leninv(nσ), nσ(n) ≠ 0, ⌜n⌝ < # nσ, stutf(nσ), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))

• Proof step 45:

  # nσ = ∞, n1 + 1 = nσ(n), ⌜nσ(n)⌝ ≠ leninv(nσ), nσ(n) ≠ 0, ⌜n⌝ < # nσ, stutf(nσ), ∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  n1 + 1 = nσ(n), ⌜nσ(n)⌝ = leninv(nσ), nσ(n) ≠ 0, ⌜n⌝ < # nσ, stutf(nσ), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦ # nσ = ∞ ∧ (n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))

Rule argument

Additional information

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

User applied rule subst equation

And obtained the following new goals

• Proof step 21:

  ⌜n1 + 1⌝ = leninv(nσ), n1 + 1 = nσ(n), n1 + 1 ≠ 0, ⌜n⌝ < # nσ, stutf(nσ), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦ # nσ = ∞ ∧ (n < m ∧ ⌜m⌝ < # nσ → nσ(m) = n1 + 1)

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 22:

  ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝

• Proof step 23:

  n1 + 1 = nσ(n), ⌜n1⌝ + ⌜1⌝ = leninv(nσ), n1 + 1 ≠ 0, ⌜n⌝ < # nσ, stutf(nσ), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦ # nσ = ∞ ∧ (n < m ∧ ⌜m⌝ < # nσ → nσ(m) = n1 + 1)

Rule argument

Additional information

Used terms: 1, n1

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  ⌜n1⌝ + ⌜1⌝ = leninv(nσ), n1 + 1 = nσ(n), ⌜n⌝ < # nσ, stutf(nσ), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦ # nσ = ∞ ∧ (n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))

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)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 25:

  ⊦ 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 26:

  n1 + 1 = nσ(n), ⌜n⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ ∃ m. nσ(m) = n1 ∧ (∀ m0. m < m0 → nσ(m0) = n1 + 1);nσ((# nσ).n - 1) = n1), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦ # nσ = ∞ ∧ (n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))

Rule argument

Additional information

Used terms: nσ, n1

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

Lemma leninv-succ

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  n1 + 1 = nσ(n), ⌜n⌝ < # nσ, stutf(nσ), (# nσ = ∞ ⊃ nσ(m0) = n1 ∧ (∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1);nσ((# nσ).n - 1) = n1), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦ # nσ = ∞ ∧ (n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))

Rule argument

Additional information

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

User applied rule case distinction

And obtained the following new goals

• Proof step 28:

  # nσ = ∞, n1 + 1 = nσ(n), nσ(m0) = n1, ⌜n⌝ < # nσ, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1, ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦ # nσ = ∞ ∧ (n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))

• Proof step 38:

  n1 + 1 = nσ(n), nσ((# nσ).n - 1) = n1, # nσ ≠ ∞, ⌜n⌝ < # nσ, stutf(nσ), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦ # nσ = ∞ ∧ (n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

  # nσ = ∞, nσ(m0) + 1 = nσ(n), nσ(m) ≠ nσ(n), n < m, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(n), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

Rule argument

Additional information

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

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

User applied rule all left

And obtained the following new goals

• Proof step 30:

  # nσ = ∞, nσ(m0) + 1 = nσ(n), nσ(m) ≠ nσ(n), n < m, stutf(nσ), m0 < n → nσ(n) = nσ(n), ∀ m1. m0 < m1 → nσ(m1) = nσ(n), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

Rule argument

Additional information

Used terms: n

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  # nσ = ∞, nσ(m0) + 1 = nσ(n), nσ(m) ≠ nσ(n), n < m, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(n), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

Rule argument

Additional information

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 32:

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

• Proof step 33:

  # nσ = ∞, nσ(m0) + 1 = nσ(n), nσ(m) ≠ nσ(n), n < m, stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(n), mon≤(nσ) ∧ n ≤ m ∧ ⌜m⌝ < # nσ → nσ(n) ≤ nσ(m), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

Rule argument

Additional information

Used terms: nσ, n, m

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

Lemma monle-leq

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 34:

  # nσ = ∞, nσ(m0) + 1 = nσ(n), nσ(m) ≠ nσ(n), n < m, nσ(n) < nσ(m), stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(n), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m < n → n ≠ 0; (s) lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ⌜m⌝ < ∞; (s) ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) ⊦ m < n → m ≤ n; (s)

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

User applied rule all left

And obtained the following new goals

• Proof step 35:

  # nσ = ∞, nσ(m0) + 1 = nσ(n), nσ(m) ≠ nσ(n), n < m, nσ(n) < nσ(m), stutf(nσ), m0 < m → nσ(m) = nσ(n), ∀ m1. m0 < m1 → nσ(m1) = nσ(n), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

Rule argument

Additional information

Used terms: m

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 36:

  # nσ = ∞, nσ(m0) + 1 = nσ(n), nσ(m) ≠ nσ(n), m ≤ m0, n < m, n < m0, nσ(n) < nσ(m), stutf(nσ), ∀ m1. m0 < m1 → nσ(m1) = nσ(n), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

Rule argument

Additional information

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

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 37:

  ⊦ 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⌝ < ∞; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (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 Used terms: nσ, m, m0

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

Lemma monle-leq

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

  nσ((# nσ).n - 1) + 1 = nσ(n), # nσ ≠ ∞, ⌜n⌝ < # nσ, stutf(nσ), ∀ nσ0, n. nσ((# nσ).n - 1) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

Rule argument

Additional information

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

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 40:

  # nσ = ⌜n0⌝ , nσ(n0 - 1) + 1 = nσ(n), # nσ ≠ ∞, ⌜n⌝ < # nσ, stutf(nσ), ∀ nσ0, n. nσ(n0 - 1) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

Rule argument

Additional information

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

Proof step 40 from proof eq-leninv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 41:

  # nσ = ⌜n0⌝ , nσ(n0 - 1) + 1 = nσ(n), n < n0, stutf(nσ), ∀ nσ0, n. nσ(n0 - 1) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

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)

Proof step 41 from proof eq-leninv (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 42:

  # nσ = ⌜n0⌝ , n0 = m0 + 1, nσ(m0) + 1 = nσ(n), n0 ≠ 0, n < n0, stutf(nσ), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

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: n0, m0

Proof step 42 from proof eq-leninv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 43:

  # nσ = ⌜m0⌝ + ⌜1⌝ , nσ(m0) + 1 = nσ(n), n ≤ m0, stutf(nσ), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

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 ≠ 0; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 43 from proof eq-leninv (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 44:

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

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ¬ n < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m = n + 1 → ¬ m ≤ n; (s) Used terms: nσ, n, m0

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

Lemma monle-leq

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 46:

  # nσ = ∞, n1 + 1 = nσ(n), nσ(n) ≠ 0, ⌜nσ(n)⌝ ≠ leninv(nσ), stutf(nσ), ∀ m. n < m → nσ(m) = nσ(n), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

Proof step 46 from proof eq-leninv (previous step) (back to tree)

User applied rule subst equation

And obtained the following new goals

• Proof step 47:

  # nσ = ∞, n1 + 1 = nσ(n), n1 + 1 ≠ 0, ⌜n1 + 1⌝ ≠ leninv(nσ), stutf(nσ), ∀ m. n < m → nσ(m) = n1 + 1, ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 48:

  ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝

• Proof step 49:

  # nσ = ∞, n1 + 1 = nσ(n), n1 + 1 ≠ 0, ⌜n1⌝ + ⌜1⌝ ≠ leninv(nσ), stutf(nσ), ∀ m. n < m → nσ(m) = n1 + 1, ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦

Rule argument

Additional information

Used terms: 1, n1

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

A spec-lemma

Proof step 49 from proof eq-leninv (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 50:

  # nσ = ∞, n1 + 1 = nσ(n), ⌜n1⌝ + ⌜1⌝ ≠ leninv(nσ), stutf(nσ), ∀ m. n < m → nσ(m) = nσ(n), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦

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)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 51:

  ⊦ 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 52:

  # nσ = ∞, n1 + 1 = nσ(n), stutf(nσ), (# nσ = ∞ ⊃ ∀ m. ¬(nσ(m) = n1 ∧ (∀ m0. m < m0 → nσ(m0) = n1 + 1));nσ((# nσ).n - 1) ≠ n1), ∀ m. n < m → nσ(m) = nσ(n), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦

Rule argument

Additional information

Used terms: nσ, n1

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

Lemma leninv-succ

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 53:

  # nσ = ∞, n1 + 1 = nσ(n), stutf(nσ), ∀ m. n < m → nσ(m) = nσ(n), ∀ m. nσ(m) ≠ n1 ∨ ¬(∀ m0. m < m0 → nσ(m0) = nσ(m) + 1), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))) ⊦

Rule argument

Additional information

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 54:

  ⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n < nσ(m) → (∃ m0. m0 < m ∧ nσ(m0) = n ∧ nσ(m0 + 1) = n + 1)

• Proof step 55:

  # nσ = ∞, n1 + 1 = nσ(n), stutf(nσ), ∀ m. nσ(m) ≠ n1 ∨ ¬(∀ m0. m < m0 → nσ(m0) = nσ(m) + 1), ∀ m. n < m → nσ(m) = nσ(n), ∀ nσ, n. n1 = nσ(n) ∧ nσ(n) ≠ 0 ∧ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ(n)⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ(m) = nσ(n))), stutf(nσ) ∧ ⌜n⌝ < # nσ ∧ n1 < nσ(n) → (∃ m0. m0 < n ∧ nσ(m0) = n1 ∧ nσ(m0 + 1) = n1 + 1) ⊦

Rule argument

Additional information

Used terms: nσ, n, n1

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

Lemma stutf-alltakenmax

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 56:

  # nσ = ∞, nσ(m0) + 1 = nσ(n), nσ(m0 + 1) = nσ(n), m0 < n, nσ(m0) < nσ(n), stutf(nσ), ∀ m. nσ(m) ≠ nσ(m0) ∨ ¬(∀ m0. m < m0 → nσ(m0) = nσ(m) + 1), ∀ m. n < m → nσ(m) = nσ(n), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

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) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic ⊦ m ≤ m; (s) ⊦ m < n → n ≠ 0; (s) lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat ⊦ m = n + 1 → ¬ m ≤ n; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ⌜m⌝ < ∞; (s)

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

User applied rule all left

And obtained the following new goals

• Proof step 57:

  # nσ = ∞, nσ(m0 + 1) = nσ(n), nσ(m0) + 1 = nσ(n), m0 < n, nσ(m0) < nσ(n), stutf(nσ), ∀ m. n < m → nσ(m) = nσ(n), nσ(m0) ≠ nσ(m0) ∨ ¬(∀ m1. m0 < m1 → nσ(m1) = nσ(m0) + 1), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

Rule argument

Additional information

Used terms: m0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 58:

  # nσ = ∞, nσ(m0 + 1) = nσ(n), nσ(m0) + 1 = nσ(n), nσ(m1) ≠ nσ(n), m0 < m1, m0 < n, nσ(m0) < nσ(n), stutf(nσ), ∀ m. n < m → nσ(m) = nσ(n), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(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 58 from proof eq-leninv (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 = ∞, nσ(m0) + 1 = nσ(n), nσ(m0 + 1) = nσ(n), nσ(m1) ≠ nσ(n), m0 < n, m0 < m1, nσ(m0) < nσ(n), stutf(nσ), ∀ m. n < m → nσ(m) = nσ(n), mon≤(nσ) ∧ m0 + 1 ≤ m1 ∧ ⌜m1⌝ < # nσ → nσ(m0 + 1) ≤ nσ(m1), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

Rule argument

Additional information

Used terms: nσ, m0 + 1, m1

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

Lemma monle-leq

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 61:

  # nσ = ∞, nσ(m0) + 1 = nσ(n), nσ(m0 + 1) = nσ(n), nσ(m1) ≠ nσ(n), m0 < n, m0 < m1, nσ(m0) < nσ(n), nσ(m0) < nσ(m1), nσ(n) < nσ(m1), stutf(nσ), ∀ m. n < m → nσ(m) = nσ(n), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true); ⊦ m ≠ n → (m ≤ n ↔ m < n); (s) 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.+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 lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ ⌜m⌝ < ∞; (s) ⊦ m + 1 ≤ n ↔ m < n; (s)

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

User applied rule all left

And obtained the following new goals

• Proof step 62:

  # nσ = ∞, nσ(m0 + 1) = nσ(n), nσ(m0) + 1 = nσ(n), nσ(m1) ≠ nσ(n), m0 < m1, m0 < n, nσ(n) < nσ(m1), nσ(m0) < nσ(m1), nσ(m0) < nσ(n), stutf(nσ), ∀ m. n < m → nσ(m) = nσ(n), n < m1 → nσ(m1) = nσ(n), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

Rule argument

Additional information

Used terms: m1

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 63:

  # nσ = ∞, nσ(m0) + 1 = nσ(n), nσ(m0 + 1) = nσ(n), nσ(m1) ≠ nσ(n), m0 < m1, m0 < n, m1 ≤ n, nσ(m0) < nσ(n), nσ(m0) < nσ(m1), nσ(n) < nσ(m1), stutf(nσ), ∀ m. n < m → nσ(m) = nσ(n), ∀ nσ0, n. nσ(m0) = nσ0(n) ∧ nσ0(n) ≠ 0 ∧ ⌜n⌝ < # nσ0 ∧ stutf(nσ0) → (⌜nσ0(n)⌝ = leninv(nσ0) ↔ # nσ0 = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ0 → nσ0(m) = nσ0(n))) ⊦

Rule argument

Additional information

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

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 64:

  ⊦ 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 ↔ m < n); (s) ⊦ m < n → ¬ n < m; (s) ⊦ ⌜m⌝ < ∞; (s) Used terms: nσ, m1, n

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

Lemma monle-leq