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Proof step 1 from proof bmin-two-03 (conclusion)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: bmin-two-01: ⊦ a < b → (∅ ++ a ++ b).butmin = ∅ ++ b;%min-two: ⊦ a < b → (∅ ++ a ++ b).min = a;