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Proof step 58 from proof inv-get-inverse-last (previous step)

nσ[m1 + 1] = nσ[m1] + 1, # nσ = ∞, stutf(nσ), ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ m0. nσ[m0] = nσ[m1] → ¬ m1 < m0 ⊦ nσ[m1 + m + 1] = nσ[m1] + 1

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: m1 + m + 1

Additional information

Used simplifier rules: s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: m1 + m + 1