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Proof step 11 from proof union-eq-empty (previous step)

zh0 ⊥ zh1, ∀ r. (r ∈ ∅ ↔ r ∈ zh0 ∪ zh1) ∧ (r ∈ ∅ → (∅)[r] = (zh0 ∪ zh1)[r]) ⊦ ¬ r ∈ zh0 ∧ ¬ r0 ∈ zh1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
zh0 ⊥ zh1, ∀ r1. ¬ r1 ∈ zh0 ∧ ¬ r1 ∈ zh1 ⊦ ¬ r ∈ zh0 ∧ ¬ r0 ∈ zh1

Rule argument

Additional information

Used simplifier rules: ⊦ zh ⊥ zh0 → (r ∈ zh ∪ zh0 ↔ r ∈ zh ∨ r ∈ zh0); (s)% ⊦ ¬ r ∈ ∅; (s)