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Proof step 8 from proof sort-max-max-list (previous step)

max(sort(x0)) = max(x0), max(x0) < a ⊦ x0 = [], max(sort(x0) + a ') = a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ x ≠ [] ∧ ordered≤(x) → max(x) = x.last
Proof step 10:
¬(sort(x0) + a ' ≠ [] ∧ ordered≤(sort(x0) + a ')), max(sort(x0)) = max(x0), max(x0) < a ⊦ x0 = [], max(sort(x0) + a ') = a

Rule argument

spec: inst: name: ordered-max seq: ⊦ x ≠ [] ∧ ordered≤(x) → max(x) = x.last subst: [x] → [sort(x0) + a '] flags: rewrite, current

Additional information

Used simplifier rules: last-def: ⊦ (x + a ').last = a; from specification list-last%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: sort(x0) + a '