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Proof step 27 from proof max-com (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
max(x3 + a0 ' + x) = (max(x + x3) < a0 ⊃ a0;max(x + x3)), max(x + a ' + x3) = (max(x3 + x) < a ⊃ a;max(x3 + x)), ∀ x1, x2. # x1 + # x2 < #(a + x3) + #(a0 + x) → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ ((max(x + x3) < a0 ⊃ a0;max(x + x3)) < a ⊃ a;(max(x + x3) < a0 ⊃ a0;max(x + x3))) = ((max(x3 + x) < a ⊃ a;max(x3 + x)) < a0 ⊃ a0;(max(x3 + x) < a ⊃ a;max(x3 + x))), x3 + x = [], x + x3 = []

Proof step 27 from proof max-com (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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