Is A-(BnC)=(A-B)U(A-C) a Valid Equation in Set Theory?

[a_skier]

A-(B\bigcapC)=(A-B)\bigcup(A-C)

If A-B={xlx\inA and x\notinB}
A-C={xlx\inA and x\notinC}

then (A-B)\bigcup(A-C)={xlx\inA, x\notin(B and C)

Let X=A and Y=(B\bigcapC)

X-Y={xlx\inX and x\notinY}
x\notinY
x\notin(B\bigcapC)
x\notin(B and C)

Therefore, A-(B\bigcapC)=(A-B)\bigcup(A-C).

Is this anywhere close to being right? I want to self study some rigorous math this summer but the hard part is that I don't have anything to check against to see if my answers are correct or not.

 

[mathman]

A-(B\bigcapC)=(A-B)\bigcup(A-C)

If A-B={xlx\inA and x\notinB}
A-C={xlx\inA and x\notinC}

then (A-B)\bigcup(A-C)={xlx\inA, x\notin(B and C)

Let X=A and Y=(B\bigcapC)

X-Y={xlx\inX and x\notinY}
x\notinY
x\notin(B\bigcapC)
x\notin(B and C)

Therefore, A-(B\bigcapC)=(A-B)\bigcup(A-C).

Is this anywhere close to being right? I want to self study some rigorous math this summer but the hard part is that I don't have anything to check against to see if my answers are correct or not.

(A-B)\bigcup(A-C)={xlx\inA, x\notin(B and C)
has to be justified.

also "and" and "\bigcap" are the same.

 

[Borek]

I always find it fascinating that people use latex just for a single symbol, instead of formatting whole expression.

A-(B\cap C)=(A-B)\cup(A-C)

 

[a_skier]

Ok so I worked through all the exercises up to this one. Here's what I came up with.

prove:
A−(B\capC)=(A−B)\cup(A−C)

Let S=B\capC={xlx\inB, and x\inC} (By the definition of intersection)

Thus A-S={xlx\inA and x\notinS}
={xlx\inA,x\notinB, and x\notinC} (see justification 1)

Next, let S_{1}=A-B={xlx\inA and x\notinB}
let S_{2}=A-C={xlx\inA and x\notinC}

Thus, S_{1}\cupS_{2}={xlx\inA. x\notinB, and x\notinC} (by the definition of union - see 2)

Therefore: A-S=S_{1}\cupS_{2}

Justifications:

1)A-B={xlx\inA and x\notinB}
2) Union is defined as the set of those elements which are in A, in B, or in both. Therefore if A={xlx satisfies P and x\notinC} and B={xlx satisfies Q and x\notinD} (where P and Q are arbitrary conditions and C and D are other sets) then A\cupB= {xlx satisfies P,Q, x\notinC, and x\notinD}.

What do you guys think?

 

[mathman]

Thus A-S={xlx∈ A and x∉ S}
={xlx∈ A,x∉ B, and x∉ C} (see justification 1)

Wrong: should be {xlx∈ A and (x∉ B or x∉ C)}