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Proof step 108 from proof inv-get-inverse-succ (previous step)

# nσ = ⌜m⌝ + ⌜1⌝ , max∞(λ n0. ¬ m < n0 ∧ nσ[n0] = n) = ⌜n1⌝ , n1 < m, n ≤ nσ[m], stutf(nσ) ⊦ m = 0, nσ[n1 + 1] = n + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 109:
⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)
Proof step 110:
# nσ = ⌜m⌝ + ⌜1⌝ , (¬ m < n1 ∧ nσ[n1] = n) ∧ (∀ m0. ¬ m < m0 ∧ nσ[m0] = n → ¬ n1 < m0), n1 < m, n ≤ nσ[m], stutf(nσ) ⊦ m = 0, nσ[n1 + 1] = n + 1

Rule argument

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

Additional information

Used terms: n1, λ n0. ¬ m < n0 ∧ nσ[n0] = n