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Proof step 118 from proof inv-get-inverse-between (previous step)

∀ n. ∃ n0. n < n0 ∧ nσ[n0] = nσ[m0], # nσ = ∞, nσ[m0] = nσ[n1] + 1, stutf(nσ), nσ[n1] < nσ[m1], n1 + m + 1 < ∞.n, m0 ≤ m1, nσ[m0] ≤ nσ[m1], ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[m1] = 0, nσ[n1 + m + 1] = nσ[m0], ∞.n = 0

User applied rule all left

And obtained the following new goals

Proof step 119:
∃ n0. m1 < n0 ∧ nσ[n0] = nσ[m0], # nσ = ∞, nσ[m0] = nσ[n1] + 1, stutf(nσ), nσ[n1] < nσ[m1], n1 + m + 1 < ∞.n, m0 ≤ m1, nσ[m0] ≤ nσ[m1], ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[m1] = 0, nσ[n1 + m + 1] = nσ[m0], ∞.n = 0

Rule argument

pos: left 1 terms: m1 abort: F discard: T computed: F

Additional information

Used terms: m1