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Proof step 50 from proof max-ordered (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 51:
¬ max(x2 + x0) < a, max(x2 + x0) = max(x2), ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) = max(x2) ∧ ordered≤(x) ∧ x3 ⊑ x → max(x2 + x3) = max(x2) ⊦ max(x2 + x1) = max(x2), x2 + x1 = [], x2 = [], x2 + x0 = [], max(x2 + x1) < a, max(x2) < a

Rule argument

pos: left 1

Additional information

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