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Proof step 5 from proof min-com (previous step)

x = a + x3, ∀ 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

Proof step 6:
∀ x1, x2. #(x1 + x2) < #((a + x3) + x0) → true → x1 + x2 = [] ∨ min(x1 + x2) = min(x2 + x1) ⊦ min((a ' + x3) + x0) = min(x0 + a ' + x3)

Rule argument

Additional information

Used simplifier rules: e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list