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Proof step 98 from proof inv-get-inverse-between (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 99:
# nσ = ∞, nσ[n] = nσ[m0], m0 < n, ∞.n + m + 1 < max∞(λ n0. nσ[n0] = nσ[m0] + 1).n, nσ[m0] < nσ[m1], stutf(nσ), m0 ≤ m1, m1 < n, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ n = 0, max∞(λ n0. nσ[n0] = nσ[m0] + 1).n = 0, nσ[∞.n + m + 1] = nσ[m0] + 1, nσ[m1] = 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%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1