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Proof step 71 from proof inv-get-inverse-succ (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 72:
# nσ = ∞, stutf(nσ), n ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, ∃ n1. m + 1 < n1 ∧ nσ[n1] = n ⊦ nσ[∞.n + 1] = n + 1

Rule argument

pos: left 5 terms: m + 1 abort: F discard: T computed: F

Additional information

Used terms: m + 1