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Proof step 59 from proof inv-get-inverse-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 60:
leninv(nσ) = ⌜n1⌝ + ⌜1⌝ , stutf(nσ), ⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n⌝ + ⌜1⌝ < # nσ, n ≤ n1 ⊦ nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + 1] = n + 1

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; 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%succ: ⊦ m +1 = m + 1; from specification nat-basic2