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Proof step 58 from proof inv-get-inverse (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 59:
∃ n1. m < n1 ∧ nσ[n1] = n, # nσ = ∞, n ≤ nσ[m], stutf(nσ), ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

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

Additional information

Used terms: m