Is it true that for all sets A and B, Ƥ(A ∩ B) = Ƥ(A) ∩ Ƥ(B) ?

I think so, and here is the proof:

  Let X ∈ Ƥ(A ∩ B)
  We will show that X ∈ Ƥ(A) ∩ Ƥ(B)

    From X ∈ Ƥ(A ∩ B), we get X ⊆ A ∩ B i.e., X ⊆ A and X ⊆ B
    It follows that X ∈ Ƥ(A) and X ∈ Ƥ(B), i.e., X ∈ Ƥ(A) ∩ Ƥ(B)
    Thus Ƥ(A ∩ B) ⇒ Ƥ(A) ∩ Ƥ(B)

  Let Y ∈ Ƥ(A) ∩ Ƥ(B)
  We will show that Y ∈ Ƥ(A ∩ B)

    From Y ∈ Ƥ(A) ∩ Ƥ(B), we get Y ∈ Ƥ(A) and Y ∈ Ƥ(B), i.e., Y ⊆ A and Y ⊆ B
    Let m ∈ Y.  Then m ∈ A and m ∈ B, i.e., m ∈ A ∩ B
    So Y ⊆ A ∩ B
    It follows that Y ∈ Ƥ(A ∩ B)
    Thus Ƥ(A) ∩ Ƥ(B) ⇒ Ƥ(A ∩ B)

# posted by rot13(Unafba Pune) @ 9:31 PM