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Proof step 1 from proof bmin-two-02 (conclusion)

⊦ a < b → (∅ ++ b ++ a).butmin = ∅ ++ b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a < b ⊦ (∅ ++ b ++ a).butmin = ∅ ++ b

Rule argument

Additional information