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Proof step 60 from proof inv-inverts (previous step)

nσ[n0] = nσ[n], nσ[n + 1] = nσ[n] + 1, stutf(nσ), ⌜n⌝ + ⌜1⌝ < # nσ, ⌜n0⌝ < # nσ, n0 < n, ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = nσ[n] → ¬ n0 < m ⊦ n0 = n, (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⌝ )[nσ[n]] = n, n = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: n

Additional information

Used simplifier rules: lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2; (forward) from specification nat-inf%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%ls-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; from specification nat-inf%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%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 Used terms: n