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Proof step 2 from proof bmin-del-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min;%min-del-01: ⊦ s -- a ≠ ∅ ∧ a ≠ s.min → (s -- a).min = s.min;%bmin-del-02: ⊦ s ≠ ∅ → s -- a -- s.min = s.butmin -- a;% ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; (ws)