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Proof step 31 from proof max-ordered (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
max((a ' + x2) + x0) = (max(x2) < a ⊃ a;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) < a ⊃ a;max(x2)), x2 + x1 = [], x2 = [], max(x2 + x1) < a

Proof step 31 from proof max-ordered (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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