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Proof step 8 from proof inv-get-inverse-last (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
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

Rule argument

spec: nat-inf-basic inst: name: elim seq: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ) subst: [n∞, m] → [leninv(nσ), m0] flags:

Additional information

Used simplifier rules: add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%sum-notinf: ⊦ n∞ = ⌜m⌝ + ⌜n⌝ → (n∞ = ∞ ↔ false); from specification nat-inf% ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: leninv(nσ), m0