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Proof step 3 from proof max-append (previous step)

∀ x0. max(x1 + x0) = max(x1) ∨ max(x1 + x0) = max(x0) ∨ x1 + x0 = [] ⊦ max((a + x1) + x0) = max(a + x1), max((a + x1) + x0) = max(x0), (a + x1) + x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ x0. max(x1 + x0) = max(x1) ∨ max(x1 + x0) = max(x0) ∨ x1 + x0 = [] ⊦ max((a ' + x1) + x0) = max(a ' + x1), max((a ' + x1) + x0) = max(x0)

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list