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Proof step 120 from proof inv-get-inverse-between (previous step)

nσ[n] = nσ[m0], # nσ = ∞, nσ[m0] = nσ[n1] + 1, m1 < n, stutf(nσ), nσ[n1] < nσ[m1], n1 + m + 1 < ∞.n, m0 ≤ m1, nσ[m0] ≤ nσ[m1], m0 < n, ∀ 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, n = 0

User applied rule all left

And closed the goal!

Rule argument

pos: left 11 terms: n abort: F discard: T computed: T

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eqs: ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1; (forward) from specification nat-basic2%eq: ⊦ m = n + 1 → ¬ m ≤ n; from specification nat Used terms: n