Prove que A\(B∩C) = (A\B)∩(A\C)
Answers
need to prove the sets are equal, that is, prove the LH side is a subset of the RH side, and vice versa.
-A×(B∩C)A×(B∩C) is the set of all ordered pairs with first entry an element of AA, and second entry an element of (B∩C)(B∩C).
-(A×B)(A×B) is the set of all ordered pairs with first entry an element of AA, and second entry an element of BB.
-(A×C)(A×C) is the set of all ordered pairs with first entry an element of AA, and second entry an element of BB.
So (A×B)∩(A×C)(A×B)∩(A×C) is the set of ordered pairs that are both in (A×B)(A×B) and (A×C)(A×C).
You need to "element chase"; that is, suppose (x,y)∈A×(B∩C)(x,y)∈A×(B∩C) and then show you must have (x,y)∈(A×B)∩(A×C)(x,y)∈(A×B)∩(A×C), then do the other direction as well. I'll do one direction for you and you can try the other yourself.
Suppose
(x,y)∈A×(B∩C)(x,y)∈A×(B∩C). Then x∈Ax∈A and y∈B∩Cy∈B∩C, meaning y∈By∈B and y∈Cy∈C. Because y∈By∈B, (x,y)∈A×B(x,y)∈A×B, and because y∈Cy∈C, (x,y)∈A×C(x,y)∈A×C. Therefore (x,y)∈(A×B)∩(A×C)(x,y)∈(A×B)∩(A×C). This tells us A×(B∩C)⊆(A×B)∩(A×C)
Proof
A×(B∪C)={(x,y):x∈A,y∈B∪C}={(x,y):x∈A,y∈B or x∈A,y∈C}={(x,y):(x,y)∈A×B or (x,y)∈A×C}=(A×B)∪(A×C)A×(B∪C)={(x,y):x∈A,y∈B∪C}={(x,y):x∈A,y∈B or x∈A,y∈C}={(x,y):(x,y)∈A×B or (x,y)∈A×C}=(A×B)∪(A×C)
Therefore, A×(B∪C)=(A×B)∪(A×C)A×(B∪C)=(A×B)∪(A×C)
answer of your question