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Proof step 100 from proof inv-ls-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 101:
leninv(nσ) = ∞, max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1) = ⌜n0⌝ , stutf(nσ), n0 ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n ⊦

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf% ⊦ ⌜m⌝ ≠ ∞; (s)% ⊦ ⌜n⌝ + n∞1 = ∞ ↔ n∞1 = ∞; (s)% ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s)% ⊦ ⌜m⌝ .n = m; (s)% ⊦ ⌜nnf⌝ [n] = nnf(n); (s)