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Proof step 3 from proof min-com (previous step)

true, ∀ x1, x2. #(x1 + x2) < #(x + x0) → true → x1 + x2 = [] ∨ min(x1 + x2) = min(x2 + x1) ⊦ x + x0 = [] ∨ min(x + x0) = min(x0 + x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information