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Proof step 34 from proof inv-get-inverse-last (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: m1 abort: F discard: T computed: F

Additional information

Used simplifier rules: ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used terms: m1