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Proof step 9 from proof inv-get-inverse-last (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
leninv(nσ) = ⌜1⌝ + ⌜n⌝ , ⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n⌝ + ⌜m⌝ + ⌜1⌝ < # nσ, stutf(nσ) ⊦ nσ[max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n + m + 1] = n + 1

Rule argument

Additional information

Used simplifier rules: add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%get-infto: m < n ⊦ (nnf to n)[m] = nnf(m); from specification stream[natstream]%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%eq-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; from specification nat-inf%sum-notinf: ⊦ n∞ = ⌜m⌝ + ⌜n⌝ → (n∞ = ∞ ↔ false); from specification nat-inf%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