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Proof step 4 from proof max-ordered (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
max((a ' + x2) + x0) = max(a ' + x2), ordered≤(x0), x1 ⊑ x0, ∀ x0, x1. max(x2 + x0) = max(x2) ∧ ordered≤(x0) ∧ x1 ⊑ x0 → max(x2 + x1) = max(x2) ∨ x2 = [] ⊦ max((a ' + x2) + x1) = max(a ' + x2)

Rule argument

Additional information

Used simplifier rules: ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)