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Proof step 1 from proof min-del-01 (conclusion)

⊦ s -- a ≠ ∅ ∧ a ≠ s.min → (s -- a).min = s.min

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: min-del: ⊦ s ≠ ∅ ∧ a ≠ s.min → (s -- a).min = s.min;%del-e-02: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; from specification set-union