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Proof step 3 from proof intu-empty (previous step)

∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ∪ s1 ⊦ (∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0) ∧ (∀ a. ¬ a ∈ s ∨ ¬ a ∈ s1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information