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Proof step 98 from proof inv-get-inverse-succ (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 99:
⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))
Proof step 100:
# nσ = ⌜m⌝ + ⌜1⌝ , ∞.n < m, stutf(nσ), n ≤ nσ[m], (∀ n0. ∃ n1. n0 < n1 ∧ ¬ m < n1 ∧ nσ[n1] = n) ∨ (∀ n0. ¬(¬ m < n0 ∧ nσ[n0] = n)) ⊦ nσ[∞.n + 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. ¬ m < n0 ∧ nσ[n0] = n] flags: <all paths>

Additional information

Used terms: λ n0. ¬ m < n0 ∧ nσ[n0] = n