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Proof step 10 from proof max-com (previous step)

x0 = a0 + x, ∀ x1, x2. # x1 + # x2 < #(a + x3) + # x0 → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ (max(x3 + x0) < a ⊃ a;max(x3 + x0)) = max(x0 + a ' + x3), x3 + x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ x1, x2. # x1 + # x2 < #(a + x3) + #(a0 + x) → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ (max(x3 + a0 ' + x) < a ⊃ a;max(x3 + a0 ' + x)) = max((a0 ' + x) + a ' + x3)

Rule argument

Additional information

Used simplifier rules: e-03: ⊦ x + a ' + y ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list