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Proof step 31 from proof inv-get-inverse-last (previous step)

# nσ = ∞, max∞(λ n. nσ[n] = nσ[m1]) = ∞, nσ[m1 + 1] = nσ[m1] + 1, 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: 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%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%l-02: ⊦ ¬ m + n < m; from specification nat-basic2 Used terms: m1 + 1