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Proof step 19 from proof inv-inverts-last (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)
Proof step 21:
# nσ = ⌜n⌝ + ⌜1⌝ , nσ[n1] = nσ[n], leninv(nσ) = ∞, stutf(nσ), ∀ m. ¬ n < m ∧ nσ[m] = nσ[n] → ¬ n1 < m ⊦ n1 = n, n < n1

Rule argument

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

Additional information

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