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Proof step 104 from proof inv-get-inverse-between (previous step)

∀ n0. nσ[n0] ≠ nσ[m0], # nσ = ∞, m0 ≤ m1, stutf(nσ), nσ[m0] < nσ[m1], ∞.n + m + 1 < max∞(λ n0. nσ[n0] = nσ[m0] + 1).n, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ nσ[m1] = 0, nσ[∞.n + m + 1] = nσ[m0] + 1, max∞(λ n0. nσ[n0] = nσ[m0] + 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