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Proof step 1 from proof intu-empty (conclusion)

⊦ s ∩ (s0 ∪ s1) = ∅ ↔ s ∩ s0 = ∅ ∧ s ∩ s1 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ∪ s1) ↔ (∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0) ∧ (∀ a. ¬ a ∈ s ∨ ¬ a ∈ s1)

Rule argument

Additional information

Used simplifier rules: int-rax: ⊦ a ∈ s1 ∩ s2 ↔ ¬ ¬(a ∈ s1 ∧ a ∈ s2);%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;%In-empty: ⊦ ¬ a ∈ ∅; from specification set-basic%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic