# $A × (B-C) = (A × B)-(A × C)$. Is the statement true or false?

If it's true, prove it, if not give specific sets $$A,B,C,D.$$

My try: I tried to work with some sets and the statement turned out to be true. But when Im using identities, I do not know how to manipulate the RHS

• for the LHS: $$( A \times B ) \cap ( A \times C' )$$
• but RHS: ?

Any help would be greatly appreciated ! thank you

• Typesetting hint: you probably want \setminus instead of / so A \setminus B gives $A \setminus B$ Commented Jul 22, 2019 at 3:50

We want to prove, that $$A\times (B-C)=(A\times B)-(A\times C)$$

So let $$(x,y)\in A\times (B-C)$$. Then $$x\in A$$ and $$y\in B$$ and $$y\notin C$$. Hence $$(x,y)\in A\times B$$ and $$(x,y)\notin A\times C$$ and we conclude $$(x,y)\in (A\times B)-(A\times C)$$.

Go through this proof one by one. If you struggle with one point, go and look at the definition of the symbols.

Now let $$(x,y)\in (A\times B)-(A\times C)$$. So $$(x,y)\in (A\times B)$$ and $$(x,y)\notin (A\times C)$$.

$$(x,y)\in A\times B$$. Hence $$x\in A$$ and $$y\in B$$.

Since $$(x,y)\notin A\times C$$ and we know, that $$x\in A$$ we have that $$y\notin C$$.

So $$y\in B-C$$ and we conclude $$(x,y)\in A\times (B-C)$$.

Indeed both sets are equal.