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Proof step 161 from proof inv-get-inverse-between (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: m0 abort: F discard: T computed: F

Additional information

Used terms: m0