I'm not very confident in what I'm doing here yet so I'd just like to get some verification that if the logic is correct or not. Thanks.
Prove of disprove:
P(R ∪ S) ⊆ P(R) ∪ P(S) ∪ P(R ∩ S)
My attempt:
Counterexample:
Let R= {1,2} and S= {2,3}
R ∪ S = {1,2,3} and R ∩ S = {2}
Now {1,3} ⊆ R ∪ S = {1,3} ∈ P(R ∪ S)
but {1,3} ∉ P(R) ∪ P(S) ∪ P(R ∩ S)
Since {1,3} ∈ P(R ∪ S) but {1,3} ∉ P(R) ∪ P(S) ∪ P(R ∩ S)
P(R ∪ S) ⊄ P(R) ∪ P(S) ∪ P(R ∩ S)
