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Proof step 40 from proof max-ordered (previous step)

max(x2 + x0) < a, max(x2) < a, ordered≤(x0), x1 ⊑ x0, a < max(x2 + x1), max(x2) < max(x2 + x1), max(x2 + x0) < max(x2 + x1), max(x2 + x0) = max(x2) ∧ ordered≤(x0) ∧ x1 ⊑ x0 → max(x2 + x1) = max(x2) ⊦ x2 + x0 = [], max(x2 + x1) = a, x2 + x1 = [], x2 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 41:
max(x2 + x0) < max(x2 + x1), max(x2) < max(x2 + x1), a < max(x2 + x1), x1 ⊑ x0, ordered≤(x0), max(x2) < a, max(x2 + x0) < a ⊦ x2 = [], x2 + x1 = [], max(x2 + x1) = a, x2 + x0 = [], max(x2 + x1) = max(x2), max(x2 + x0) = max(x2)

Rule argument

Additional information

Used simplifier rules: irreflexivity: ⊦ ¬ a < a; from specification oelem%lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem