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Proof step 2 from proof union-eq-empty (previous step)

zh0 ⊥ zh1 ⊦ ∅ = zh0 ∪ zh1 ↔ zh0 = ∅ ∧ zh1 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∅ = zh0 ∪ zh1, zh0 ⊥ zh1 ⊦ zh0 = ∅ ∧ zh1 = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ ∅ ∪ zh = zh; (s)