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Proof step 9 from proof max-com (previous step)

∀ 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 = []

User applied rule constructor cut

And obtained the following new goals

Proof step 10:
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 = []

Rule argument

var: x0

Additional information

Used simplifier rules: max-rec: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x));%list freely generated by [], +;;%app: ⊦ x + [] = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%Nil: ⊦ [] + x = x; from specification list