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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
(max(x2 + x0) < a ⊃ a;max(x2 + x0)) = max(x2), max(x2 + x1) < a, ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) = max(x2) ∧ ordered≤(x) ∧ x3 ⊑ x → max(x2 + x3) = max(x2) ⊦ x2 + x0 = [], x2 + x1 = [], a = max(x2), x2 = []

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information