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Proof step 94 from proof inv-get-inverse-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 95:
# nσ = ⌜m⌝ + ⌜1⌝ , stutf(nσ), n ≤ nσ[m], max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = n).n < m ⊦ nσ[max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = n).n + 1] = n + 1, m = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%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