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Proof step 15 from proof max-com (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
max(x3 + a0 ' + x) = max((a0 ' + x) + x3), ∀ x1, x2. # x1 + # x2 < #(a + x3) + #(a0 + x) → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ (max((a0 ' + x) + x3) < a ⊃ a;max((a0 ' + x) + x3)) = (max(x + a ' + x3) < a0 ⊃ a0;max(x + a ' + x3))

Rule argument

Additional information

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