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Proof step 89 from proof inv-ls-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 90:
# nσ = ⌜n0⌝ , nσ[m0] = n + 1, m < n0, m0 < n0, n < nσ[m], stutf(nσ), ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, m0 ≤ m, ∀ n0. ¬ ⌜n0⌝ < # nσ ∨ nσ[n0] ≠ n + 1, ∀ m0. (∃ m1. m1 < n0 ∧ nσ[m1] = m0 - 1) → ¬ nσ[m] + 1 < m0 ⊦

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% ⊦ ⌜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% ⊦ m + 1 ≤ n ↔ m < n; (s)