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Proof step 34 from proof max-ordered (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
max((a ' + x2) + x0) = a, max(x2) < max(x2 + x1), a < max(x2 + x1), max(x2) < a, ordered≤(x0), x1 ⊑ x0, ∀ x, x3. max(x2 + x) = max(x2) ∧ ordered≤(x) ∧ x3 ⊑ x → max(x2 + x3) = max(x2) ⊦ x2 + x1 = [], max(x2 + x1) = a, x2 = []

Rule argument

Additional information

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