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Proof step 42 from proof inv-get-inverse-between (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 43:
# nσ = ∞, nσ[m0] = n + 1, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , m0 < m1, n + 1 < nσ[m1], n1 + m + 1 < ∞.n, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ m. ∃ m0. m < nσ[m0], ∃ n1. m1 < n1 ∧ nσ[n1] = n + 1 ⊦ nσ[n1 + m + 1] = n + 1

Rule argument

pos: left 9 terms: m1 abort: F discard: T computed: F

Additional information

Used terms: m1