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Proof step 72 from proof inv-ls-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 73:
nσ[m] = n + 1, nσ[n0] = n + 1, stutf(nσ), ⌜m⌝ < # nσ, ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, m < m0, m < n0, ⌜n0⌝ < # nσ, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1, ∀ m2. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m2 - 1))) → ¬ 2 + n < m2, ⌜m0⌝ < # nσ → n < nσ[m0] ⊦ nσ[m0] = n + 1

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (¬ mon≤(nσ) ↔ false);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%lels-03: ⊦ m < n → m ≤ n; from specification nat