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Proof step 45 from proof inv-ls-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
leninv(nσ) = ⌜n1⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1) = ∞, 1 + n < n1, ∞.n ≤ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[n], stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%get-infto: m < n ⊦ (nnf to n)[m] = nnf(m); from specification stream[natstream]%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%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%ls-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; from specification nat-inf%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic