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Proof step 5 from proof leninv-zero (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)
Proof step 7:
stutf(nσ) ⊦ (0 ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = 0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = 0 - 1)))) ∧ (∀ m. (m ≠ 0 → (∃ m0. ⌜m0⌝ < # nσ ∧ nσ[m0] = m - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m1. m0 < m1 → nσ[m1] = m - 1)))) → ¬ 0 < m) ↔ nσ = ⌜λ n. 0⌝

Rule argument

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

Additional information

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