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Proof step 20 from proof max-ordered (previous step)

max(a ' + x0) = a, a < max(x1), ordered≤(x0), x1 ⊑ x0 ⊦ max(x1) = a, x1 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 22:
(max(x0) < a ⊃ a;max(x0)) = a, a < max(x1), ordered≤(x0), x1 ⊑ x0 ⊦ max(x1) = a, x0 = [], x1 = []

Rule argument

spec: <toplevel> name: max-rec seq: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)) subst: [x, a] → [x0, a] flags: <all paths>

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a;% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x ⊑ [] ↔ x = []; (s) Used terms: x0, a