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Proof step 104 from proof inv-ls-succ (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 105:
⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))
Proof step 106:
(∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n) ∨ (∀ n0. ¬(⌜n0⌝ < # nσ ∧ nσ[n0] = n)), nσ[n0] = n + 1, leninv(nσ) = ∞, ⌜n0⌝ < # nσ, stutf(nσ), n0 ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, ∀ m. ⌜m⌝ < # nσ ∧ nσ[m] = n + 1 → ¬ n0 < m ⊦

Rule argument

spec: inst: name: maxinf-inf seq: ⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n)) subst: [np] → [λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n] flags: rewrite, current

Additional information

Used terms: λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n