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Proof step 26 from proof inv-get-inverse-last (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 27:
# nσ = ∞, nσ[n0] = nσ[m1] + 1, max∞(λ n. nσ[n] = nσ[m1]) = ∞, m1 < n0, stutf(nσ), m1 < n0 → nσ[n0] = nσ[n0], ∀ n1. ¬ n1 < n0 → nσ[n1] = nσ[m1], ∀ m0. m1 < m0 → nσ[m0] = nσ[n0] ⊦ nσ[∞.n + m + 1] = nσ[n0]

Rule argument

pos: left 6 terms: n0 abort: F discard: F computed: T

Additional information

Used terms: n0