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Proof step 23 from proof inv-inverts-last (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 24:
leninv(nσ) = ⌜n1⌝ , # nσ = ⌜n⌝ + ⌜1⌝ , stutf(nσ) ⊦ leninv(nσ) = ∞, ((λ m. max∞(λ n0. ¬ n < n0 ∧ nσ[n0] = m).n) to n1)[nσ[n]] = n

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: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: leninv(nσ), m0