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Proof step 13 from proof inv-get-inverse-between (previous step)

max∞(λ n0. nσ[n0] = n) = ∞, # nσ = ∞, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ max∞(λ n0. nσ[n0] = n + 1).n = 0, nσ[∞.n + m + 1] = n + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))
Proof step 15:
(∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n) ∨ (∀ n0. nσ[n0] ≠ n), # nσ = ∞, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∞.n + m + 1 < max∞(λ n0. nσ[n0] = n + 1).n, ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ max∞(λ n0. nσ[n0] = n + 1).n = 0, nσ[∞.n + m + 1] = n + 1

Rule argument

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

Additional information

Used terms: λ n0. nσ[n0] = n