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Proof step 27 from proof inv-get-inverse-last (previous step)

nσ[n0] = nσ[m1] + 1, max∞(λ n. nσ[n] = nσ[m1]) = ∞, # nσ = ∞, stutf(nσ), m1 < n0, ∀ n1. ¬ n1 < n0 → nσ[n1] = nσ[m1], ∀ m0. m1 < m0 → nσ[m0] = nσ[n0] ⊦ nσ[∞.n + m + 1] = nσ[n0]

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: n0 abort: F discard: F computed: T

Additional information

Used simplifier rules: ref-1: ⊦ ¬ n < n; from specification nat-basic1%eq: ⊦ m = m + n ↔ n = 0; 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: n0