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Proof step 8 from proof union-eq-empty (previous step)

∅ = zh0 ∪ zh1, zh0 ⊥ zh1 ⊦ ¬ r ∈ zh0 ∧ (∀ r. (r ∈ zh1 ↔ r ∈ ∅) ∧ (r ∈ zh1 → zh1[r] = (∅)[r]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∅ = zh0 ∪ zh1, zh0 ⊥ zh1 ⊦ ¬ r ∈ zh0 ∧ ¬ r0 ∈ zh1

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ r ∈ ∅; (s)