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Proof step 180 from proof inv-get-inverse-between (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 181:
⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)
Proof step 182:
# nσ = ⌜m0⌝ + ⌜1⌝ , (¬ m0 < n ∧ nσ[n] = nσ[n1] + 1) ∧ (∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] + 1 → ¬ n < m), n1 ≤ m0, stutf(nσ), nσ[n1] < nσ[m0], n1 + m + 1 < n, ∀ m. ¬ m0 < m ∧ nσ[m] = nσ[n1] → ¬ n1 < m ⊦ nσ[m0] = 0, nσ[n1 + m + 1] = nσ[n1] + 1, n = 0

Rule argument

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

Additional information

Used terms: n, λ n0. ¬ m0 < n0 ∧ nσ[n0] = nσ[n1] + 1