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Proof step 88 from proof inv-ls-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 89:
# nσ = ⌜n0⌝ , nσ[m0] = n + 1, ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, stutf(nσ), m < n0, m0 ≤ m, m0 < n0, n < nσ[m], ∀ n0. ¬ ⌜n0⌝ < # nσ ∨ nσ[n0] ≠ n + 1, ∀ m1. (∃ m0. m0 < n0 ∧ nσ[m0] = m1 - 1) → ¬ nσ[m] + 1 < m1 ⊦

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%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