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Proof step 20 from proof inv-get-inverse-between (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 21:
# nσ = ∞, ⌜nσ[m0]⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ[n0] = nσ[m0] + 1).n, m0 < m1, nσ[m0] < nσ[m1], ∃ n1. m1 < n1 ∧ nσ[n1] = nσ[m0], ∀ m. ∃ m0. m < nσ[m0] ⊦ max∞(λ n0. nσ[n0] = nσ[m0] + 1).n = 0, nσ[∞.n + m + 1] = nσ[m0] + 1, m1 = 0, nσ[m1] = 0

Rule argument

pos: left 7 terms: m1 abort: F discard: T computed: F

Additional information

Used terms: m1