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Proof step 5 from proof max-com (previous step)

x = a + x3, ∀ x1, x2. # x1 + # x2 < # x + # x0 → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ x + x0 = [], max(x + x0) = max(x0 + x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ x1, x2. # x1 + # x2 < #(a + x3) + # x0 → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ max((a ' + x3) + x0) = max(x0 + a ' + x3)

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