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Proof step 20 from proof max-com (previous step)

¬ x3 + x ≠ [], max(x3 + a0 ' + x) = max((a0 ' + 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) ⊦ (max((a0 ' + x) + x3) < a ⊃ a;max((a0 ' + x) + x3)) = (max((a ' + x3) + x) < a0 ⊃ a0;max((a ' + x3) + x))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
∀ x1, x2. # x1 + # x2 < #(a + []) + #(a0 + []) → true → x1 + x2 = [] ∨ max(x1 + x2) = max(x2 + x1) ⊦ (a0 < a ⊃ a;a0) = (a < a0 ⊃ a0;a)

Rule argument

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a;%app: ⊦ x + [] = x; from specification list%Nil: ⊦ [] + x = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%app-02-ant: ⊦ ¬ x + y ≠ [] ↔ x = [] ∧ y = []; from specification list