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Proof step 1 from proof size-two-01 (conclusion)

⊦ #(∅ ++ a ++ b) = 2 ↔ a ≠ b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ #(∅ ++ a ++ b) = 2, a = b

Rule argument

Additional information

Used simplifier rules: add-01: ⊦ s ++ a ++ a = s ++ a;%size-one-01: ⊦ #(∅ ++ a) = 1;