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Proof step 1 from proof disjoint-union-empty (conclusion)

⊦ st0 ∪ st1 = ∅ ↔ st0 = ∅ ∧ st1 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
st0 ∪ st1 = ∅ ⊦ st0 = ∅ ∧ st1 = ∅

Rule argument

Additional information

Used simplifier rules: union-right-empty: ⊦ ∅ ∪ st = st;