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Proof step 2 from proof bmin-del-01 (previous step)

⊦ s.butmin -- a -- s.min = s.butmin -- a, s = ∅

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min;% ⊦ s -- b -- a = s -- a -- b ↔ true; (s)% ⊦ s -- a -- b -- a = s -- b -- a; (s)