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Proof step 104 from proof inv-get-inverse-succ (previous step)

∀ n0. ∃ n1. n0 < n1 ∧ ¬ m < n1 ∧ nσ[n1] = nσ[m0], # nσ = ⌜m⌝ + ⌜1⌝ , m0 ≤ m, ∞.n < m, nσ[m0] ≤ nσ[m], stutf(nσ) ⊦ m = 0, nσ[∞.n + 1] = nσ[m0] + 1

User applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: m abort: F discard: T computed: F

Additional information

Used terms: m