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Proof step 29 from proof inv-get-inverse-last (previous step)

# nσ = ∞, max∞(λ n. nσ[n] = nσ[m1]) = ∞, stutf(nσ), n0 ≤ m1, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ n1. ¬ n1 < n0 → nσ[n1] = nσ[m1] ⊦ nσ[∞.n + m + 1] = nσ[m1] + 1

User applied rule all left

And obtained the following new goals

Proof step 30:
# nσ = ∞, max∞(λ n. nσ[n] = nσ[m1]) = ∞, stutf(nσ), n0 ≤ m1, m1 < m1 + 1 → nσ[m1 + 1] = nσ[m1] + 1, ∀ n1. ¬ n1 < n0 → nσ[n1] = nσ[m1] ⊦ nσ[∞.n + m + 1] = nσ[m1] + 1

Rule argument

pos: left 5 terms: m1 + 1 abort: F discard: T computed: T

Additional information

Used terms: m1 + 1