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Proof step 76 from proof inv-ls-succ (previous step)

nσ[m] = n + 1, nσ[n0] = n + 1, ⌜m0⌝ < # nσ, stutf(nσ), ⌜m⌝ < # nσ, ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, m < m0, m < n0, ⌜n0⌝ < # nσ, n + 1 < nσ[m0], n < nσ[m0], ∃ n1. m0 < 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 ⊦ nσ[m0] = n + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 77:
nσ[n1] = n + 1, nσ[n0] = n + 1, nσ[m] = n + 1, m < n1, ⌜n1⌝ < # nσ, m0 < n1, n < nσ[m0], ⌜n0⌝ < # nσ, m < n0, m < m0, ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, ⌜m⌝ < # nσ, stutf(nσ), ⌜m0⌝ < # nσ, n + 1 < nσ[m0], m + 1 < n1, ∀ m2. (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m2 - 1))) → ¬ 2 + n < m2 ⊦ nσ[m0] = n + 1

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2%lels-03: ⊦ m < n → m ≤ n; from specification nat%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat