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Proof step 35 from proof inv-get-inverse-between (previous step)

# nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , max∞(λ n0. nσ[n0] = n + 1) = ∞, n1 + m + 1 < ∞.n, ⌜n⌝ + ⌜1⌝ < leninv(nσ), stutf(nσ), ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[n1 + m + 1] = n + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 36:
⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))
Proof step 37:
# nσ = ∞, max∞(λ n0. nσ[n0] = n) = ⌜n1⌝ , ⌜n⌝ + ⌜1⌝ < leninv(nσ), n1 + m + 1 < ∞.n, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ m. ∃ m0. m < nσ[m0], ∀ n. ∃ m. nσ[m] = n, (∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n + 1) ∨ (∀ n0. nσ[n0] ≠ n + 1) ⊦ nσ[n1 + m + 1] = n + 1

Rule argument

spec: <toplevel> name: maxinf-inf seq: ⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n)) subst: [np] → [λ n0. nσ[n0] = n + 1] flags: <all paths>

Additional information

Used terms: λ n0. nσ[n0] = n + 1