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Proof step 45 from proof inv-get-inverse-between (previous step)

∀ n0. nσ[n0] ≠ n + 1, # nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , nσ[m0] = n + 1, n1 + m + 1 < ∞.n, stutf(nσ), ⌜n⌝ + ⌜1⌝ < leninv(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦ nσ[n1 + m + 1] = n + 1, ∞.n = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: m0

Additional information

Used terms: m0