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Proof step 49 from proof inv-ls-succ (previous step)

max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1) = ∞, max∞(λ n0. n0 ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1)))) = ⌜n1⌝ , 1 + n < n1, ∞.n ≤ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[n], stutf(nσ) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 50:
⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)
Proof step 51:
max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n + 1) = ∞, (n1 ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n1 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n1 - 1)))) ∧ (∀ m. (m ≠ 0 → (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m - 1)))) → ¬ n1 < m), 1 + n < n1, ∞.n ≤ ((λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ[n] = m).n) to n1)[n], stutf(nσ) ⊦

Rule argument

spec: inst: name: maxinf-fin seq: ⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m) subst: [n, np] → [n1, λ n0. n0 ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1)))] flags: rewrite, current

Additional information

Used terms: n1, λ n0. n0 ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1)))