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Proof step 41 from proof inv-get-inverse-between (previous step)

nσ[m0] = n + 1, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , # nσ = ∞, n + 1 < nσ[m1], m0 < m1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), n1 + m + 1 < ∞.n, ∀ m. ∃ m0. m < nσ[m0], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n + 1 ⊦ nσ[m1] = 0, m1 = 0, ∞.n = 0, nσ[n1 + m + 1] = n + 1

User applied rule all left

And obtained the following new goals

Proof step 42:
nσ[m0] = n + 1, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , # nσ = ∞, n + 1 < nσ[m1], m0 < m1, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), n1 + m + 1 < ∞.n, ∀ m. ∃ m0. m < nσ[m0], ∃ n1. m1 < n1 ∧ nσ[n1] = n + 1 ⊦ nσ[m1] = 0, m1 = 0, ∞.n = 0, nσ[n1 + m + 1] = n + 1

Rule argument

pos: left 10 terms: m1 abort: F discard: T computed: F

Additional information

Used terms: m1