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Proof step 145 from proof inv-get-inverse-between (previous step)

nσ[0] = 0, nσ[n] = nσ[n1] + 1, # nσ = ∞, nσ[n1 + 1] ≤ nσ[n1 + m + 1], nσ[n] ≤ nσ[m1], n1 + m + 1 < n, nσ[n1] < nσ[m1], n ≤ m1, n1 + m + 1 < m1, n1 < m1, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ m. nσ[m] = nσ[n] → ¬ n < m, ∀ n0. nσ[n0 + 1] = nσ[n0] ∨ nσ[n0 + 1] = nσ[n0] + 1 ⊦ nσ = [], n = 0, nσ[n1 + m + 1] = nσ[n], nσ[m1] = 0, nσ[n] = nσ[m1] + 1, nσ[n1 + 1] = nσ[n1], nσ[n1 + 1] = nσ[n]

User applied rule all left

And closed the goal!

Rule argument

pos: left 14 terms: n1 abort: F discard: T computed: T

Additional information

Used terms: n1