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Proof step 112 from proof inv-get-inverse-between (previous step)

max∞(λ n0. nσ[n0] = nσ[n1] + 1) = ∞, # nσ = ∞, n1 + m + 1 < ∞.n, nσ[n1] < nσ[m1], stutf(nσ), ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ ∞.n = 0, nσ[n1 + m + 1] = nσ[n1] + 1, nσ[m1] = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 113:
⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))
Proof step 114:
(∀ n. ∃ n0. n < n0 ∧ nσ[n0] = nσ[n1] + 1) ∨ (∀ n. nσ[n] ≠ nσ[n1] + 1), # nσ = ∞, n1 + m + 1 < ∞.n, nσ[n1] < nσ[m1], stutf(nσ), ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1 ⊦ ∞.n = 0, nσ[n1 + m + 1] = nσ[n1] + 1, nσ[m1] = 0

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] = nσ[n1] + 1] flags: rewrite, current

Additional information

Used terms: λ n0. nσ[n0] = nσ[n1] + 1