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Proof step 36 from proof max-ordered (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
(max(x2 + x0) < a ⊃ a;max(x2 + x0)) = a, max(x2) < a, ordered≤(x0), x1 ⊑ x0, a < max(x2 + x1), max(x2) < max(x2 + x1), ∀ x, x3. max(x2 + x) = max(x2) ∧ ordered≤(x) ∧ x3 ⊑ x → max(x2 + x3) = max(x2) ⊦ x2 + x0 = [], max(x2 + x1) = a, x2 + x1 = [], x2 = []

Rule argument

Additional information

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