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Proof step 3 from proof union-inter-inter-union (previous step)

⊦ ∀ a. a ∈ (s0 ∪ s1) ∩ s2 ↔ a ∈ (s0 ∩ s2) ∪ (s1 ∩ s2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ s0 ∧ a ∈ s2 ∨ a ∈ s1 ∧ a ∈ s2 ⊦ (a ∈ s0 ∨ a ∈ s1) ∧ a ∈ s2

Rule argument

Additional information

Used simplifier rules: int-rax: ⊦ a ∈ s1 ∩ s2 ↔ ¬ ¬(a ∈ s1 ∧ a ∈ s2);%unrax: ⊦ a ∈ s ∪ s0 ↔ ¬ ¬(a ∈ s ∨ a ∈ s0);%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;%Intersect: ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2;