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Proof step 40 from proof inv-get-inverse-between (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 41:
# nσ = ∞, nσ[m0] = n + 1, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < ∞.n, stutf(nσ), ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n + 1, ∀ m. ∃ m0. m < nσ[m0], ∃ m1. m0 < m1 ∧ nσ[m0] < nσ[m1] ⊦ nσ[n1 + m + 1] = n + 1

Rule argument

pos: left 8 terms: m0 abort: F discard: T computed: F

Additional information

Used terms: m0