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Proof step 13 from proof union-eq-empty (previous step)

zh0 ⊥ zh1, ∀ r1. ¬ r1 ∈ zh0 ∧ ¬ r1 ∈ zh1 ⊦ r ∈ zh1, r ∈ zh0, ¬ r ∈ zh0 ∧ ¬ r0 ∈ zh1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
r0 ∈ zh1, zh0 ⊥ zh1, ∀ r1. ¬ r1 ∈ zh0 ∧ ¬ r1 ∈ zh1 ⊦ r ∈ zh0, r ∈ zh1

Rule argument

Additional information