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Proof step 21 from proof max-com (previous step)

max(x3 + a0 ' + x) = (max(x + x3) < a0 ⊃ a0;max(x + x3)), x + (a + x3) = [] ∨ max(x + (a + x3)) = max((a + x3) + x), ∀ x1, x2. # x1 + # x2 < #(a + x3) + #(a0 + x) → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ x + x3 = [], ((max(x + x3) < a0 ⊃ a0;max(x + x3)) < a ⊃ a;(max(x + x3) < a0 ⊃ a0;max(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 22:
max(x3 + a0 ' + x) = (max(x + x3) < a0 ⊃ a0;max(x + x3)), max(x + a ' + x3) = max((a ' + x3) + x), ∀ x1, x2. # x1 + # x2 < #(a + x3) + #(a0 + x) → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ x + x3 = [], ((max(x + x3) < a0 ⊃ a0;max(x + x3)) < a ⊃ a;(max(x + x3) < a0 ⊃ a0;max(x + x3))) = (max((a ' + x3) + x) < a0 ⊃ a0;max((a ' + x3) + x))

Rule argument

Additional information

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