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Proof step 12 from proof max-append (previous step)

max(x1 + x0) < a, ∀ x0. max(x1 + x0) = max(x1) ∨ max(x1 + x0) = max(x0) ∨ x1 + x0 = [] ⊦ x1 + x0 = [], a = (max(x1) < a ⊃ a;max(x1)), a = max(x0), x1 = []

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

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