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Proof step 28 from proof inv-get-inverse-last (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: n0 abort: F discard: F computed: T

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)% ⊦ m = m + n ↔ n = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n0