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Proof step 150 from proof inv-get-inverse-between (previous step)

n1 = m0 + 1 ∧ n1 ≠ 0, # nσ = ⌜n1⌝ , stutf(nσ), n < nσ[m0], max∞(λ n0. n0 < n1 ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. n0 < n1 ∧ nσ[n0] = n + 1).n ⊦ nσ[m0] = 0, nσ[max∞(λ n0. n0 < n1 ∧ nσ[n0] = n).n + m + 1] = n + 1, max∞(λ n0. n0 < n1 ∧ nσ[n0] = n + 1).n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 151:
# nσ = ⌜m0⌝ + ⌜1⌝ , stutf(nσ), n < nσ[m0], max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n).n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n ⊦ nσ[m0] = 0, nσ[max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n).n + m + 1] = n + 1, max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf