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Proof step 36 from proof inv-get-inverse-between (previous step)

(∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n + 1) ∨ (∀ n0. nσ[n0] ≠ n + 1), # nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , n1 + m + 1 < ∞.n, stutf(nσ), ⌜n⌝ + ⌜1⌝ < leninv(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ n. ∃ m. nσ[m] = n ⊦ nσ[n1 + m + 1] = n + 1, ∞.n = 0

User applied rule all left

And obtained the following new goals

Proof step 37:
(∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n + 1) ∨ (∀ n0. nσ[n0] ≠ n + 1), # nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , n1 + m + 1 < ∞.n, stutf(nσ), ⌜n⌝ + ⌜1⌝ < leninv(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∃ m. nσ[m] = n + 1 ⊦ nσ[n1 + m + 1] = n + 1, ∞.n = 0

Rule argument

pos: left 9 terms: n + 1 abort: F discard: T computed: F

Additional information

Used terms: n + 1