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Proof step 32 from proof inv-get-inverse-last (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: m1 + 1 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ 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%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ m + n < m; (s) Used terms: m1 + 1