Given nonempty sets A and B, prove that every set in P(B − A) ⊆ P(B) − P(A).

Proof:Let [tex]X \in P(B-A) [/tex]. As we chose [tex]X[/tex] in [tex]P(B-A) [/tex] we know that [tex]X \subseteq B-A[/tex]. Since [tex]B-A \subseteq B[/tex] by transitivity we get:[tex]X \subseteq B \quad \implies X \in P(B)[/tex]. If [tex]X [/tex] is the empty set, we already have that [tex]X = \emptyset \in P(B) - P(A)[/tex]. But if [tex]X[/tex] is not empty, that means that it can't be subset of [tex]A[/tex], because [tex] X [/tex] is already  subset of [tex]B-A[/tex], and those sets do not share any element. In other words:[tex]X \subseteq A \cup (B-A) = \emptyset[/tex][tex] \Rightarrow X = \emptyset[/tex]As [tex]X[/tex] can't be subset of [tex]A[/tex], then [tex] X\notin P(A) [/tex]. [tex]X[/tex] was an arbitrary element, and [tex]X \in P(B)[/tex][tex] X\notin P(A) [/tex]Thus, [tex] X\in P(B)-P(A) [/tex], where we conclude that [tex]P(B-A) \subseteq P(B) - P(A)[/tex]