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Proof step 14 from proof max-ordered (previous step)

(max(x2 + x0) < a ⊃ a;max(x2 + x0)) = max(x2), x1 ⊑ x0, ordered≤(x0), max(x2 + x1) < a, ∀ x, x3. max(x2 + x) = max(x2) ∧ ordered≤(x) ∧ x3 ⊑ x → max(x2 + x3) = max(x2) ⊦ x2 = [], x2 + x1 = [], a = max(x2), x2 + x0 = []

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: max-ge: ⊦ x ≠ [] → (¬ ¬ max(x + x0) < max(x) ↔ false);%sls: ⊦ a ≠ b → (¬ a < b ↔ b < a); from specification oelem%transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem%a: ⊦ (x + y) + z = x + y + z; from specification list