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Proof step 65 from proof inv-get-inverse-between (previous step)

nσ[n1] = nσ[n1 + m + 1], nσ[m1 + 1] = nσ[n1] + 1, # nσ = ∞, stutf(nσ), ⌜nσ[n1]⌝ + ⌜1⌝ < leninv(nσ), nσ[m1] ≤ nσ[n1] + 1, n1 + m < m1, nσ[n1] ≤ nσ[m1], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ n. ∃ m. nσ[m] = n, ∀ m. nσ[m] = nσ[n1] + 1 → ¬ m1 + 1 < m, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m ⊦ m1 = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 13 terms: n1 + 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: n1 + m + 1