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Proof step 70 from proof inv-ls-succ (previous step)

nσ[n0] = n + 1, n < nσ[m], ⌜m⌝ < # nσ, m < m0, ⌜n0⌝ < # nσ, m < n0, stutf(nσ), ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, mon≤(nσ) ∧ m ≤ n0 ∧ ⌜n0⌝ < # nσ → nσ[m] ≤ nσ[n0], ∀ m0. (∃ m1. ⌜m1⌝ < # nσ ∧ nσ[m1] = m0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m3. m1 < m3 → nσ[m3] = m0 - 1))) → ¬ nσ[m] + 1 < m0, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1 ⊦ nσ[m0] = nσ[m]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 71:
nσ[n0] = n + 1, nσ[m] = n + 1, ⌜m⌝ < # nσ, m < m0, ⌜n0⌝ < # nσ, m < n0, stutf(nσ), ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, ∀ m2. (∃ m1. ⌜m1⌝ < # nσ ∧ nσ[m1] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m3. m1 < m3 → nσ[m3] = m2 - 1))) → ¬ 2 + n < m2, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1 ⊦ nσ[m0] = n + 1

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true);% ⊦ m + 1 ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (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 → (n ≤ m + 1 ↔ n = m + 1); (s)% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m ≤ m; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat% ⊦ m < n → m ≤ n; (s)