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Proof step 1 from proof bmin-sub (conclusion)

⊦ s ⊆ s0 ∧ s ≠ ∅ → s.butmin ⊆ s0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min;%sub-del-03: ⊦ s ⊆ s0 → s -- a ⊆ s0; from specification set-union