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Proof step 73 from proof inv-get-inverse-succ (previous step)

# nσ = ∞, m + 1 < n0, stutf(nσ), nσ[n0] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ nσ[∞.n + 1] = nσ[n0] + 1

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n0 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ ¬ m < n → ¬ m + m0 < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m = n + 1 → ¬ m ≤ n; (s) Used terms: n0