# Proof of A-(BnC)=(A-B)U(A-C)

1. Jun 16, 2013

### a_skier

A-(B$\bigcap$C)=(A-B)$\bigcup$(A-C)

If A-B={xlx$\in$A and x$\notin$B}
A-C={xlx$\in$A and x$\notin$C}

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

Let X=A and Y=(B$\bigcap$C)

X-Y={xlx$\in$X and x$\notin$Y}
x$\notin$Y
x$\notin$(B$\bigcap$C)
x$\notin$(B and C)

Therefore, A-(B$\bigcap$C)=(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.

2. Jun 17, 2013

### mathman

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

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

3. Jun 17, 2013

### Staff: Mentor

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

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

4. Jun 18, 2013

### a_skier

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

prove:
A−(B$\cap$C)=(A−B)$\cup$(A−C)

Let S=B$\cap$C={xlx$\in$B, and x$\in$C} (By the definition of intersection)

Thus A-S={xlx$\in$A and x$\notin$S}
={xlx$\in$A,x$\notin$B, and x$\notin$C} (see justification 1)

Next, let S$_{1}$=A-B={xlx$\in$A and x$\notin$B}
let S$_{2}$=A-C={xlx$\in$A and x$\notin$C}

Thus, S$_{1}$$\cup$S$_{2}$={xlx$\in$A. x$\notin$B, and x$\notin$C} (by the definition of union - see 2)

Therefore: A-S=S$_{1}$$\cup$S$_{2}$

Justifications:

1)A-B={xlx$\in$A and x$\notin$B}
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$\notin$C} and B={xlx satisfies Q and x$\notin$D} (where P and Q are arbitrary conditions and C and D are other sets) then A$\cup$B= {xlx satisfies P,Q, x$\notin$C, and x$\notin$D}.

What do you guys think?

5. Jun 19, 2013

### mathman

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