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Proof step 119 from proof inv-get-inverse-between (previous step)

∃ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 120:
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

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat