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Proof step 40 from proof inv-get-inverse-last (previous step)

# nσ = ⌜m0⌝ + ⌜1⌝ , max∞(λ n. ¬ m0 < n ∧ nσ[n] = nσ[m0]) = ∞, m + ∞.n < m0, stutf(nσ), ∀ n. ¬ m0 < n → nσ[n] < nσ[m0] ⊦ nσ[∞.n + m + 1] = nσ[m0] + 1

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: m0

Additional information

Used simplifier rules: ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used terms: m0