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Result: $(A\times C)-(B\times C)\subseteq (A-B)\times C$

Proof: Let $(x,y)\in (A\times C)-(B\times C)$. Then $(x,y)\in A\times C,\implies x\in A,y\in C$. Since $(x,y)\notin B\times C, x\notin B$. Thus $x\in A-B$ and hence $(x,y)\in (A-B)\times C$

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Nothing wrong with it. Why do you think there is? –  Patrick Da Silva Oct 20 '12 at 17:58
    
As this is a question which required to evaluate the proposed proof of the result and i am not sure how to do it –  Mathematics Oct 20 '12 at 17:59
    
Good proof, brief too. Grammatically a bit funny. For example, "Let $X$ $\implies$ $Y$" is non-standard English syntax, understandable but ungrammatical. And some sentences do not end with periods. –  André Nicolas Oct 20 '12 at 18:06
    
And the logic is not entirely clear. Overuse of $\implies$. Also, should separate what $(x,y)\in A\times C$ implies from what $(x,y)\not\in B\times C$ implies. –  André Nicolas Oct 20 '12 at 18:13
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1 Answer 1

The proof is in principle correct. However, one might add that $(x,y)\notin B\times C$ gives you immediately only $x\notin B\lor y\notin C$. But together with $y\in C$ (from $(x,y)\in A\times C$), you get indeed that $x\notin B$.

If one really wants to make that proof very explicit, one may werite:

Let $z\in (A\times C)-(B\times C)$ be an arbitrary element. Then $z\in A\times C$ and $z\notin B\times C$, hence there exist $x\in A$, $y\in C$ with $z=(x,y)$. If we assume $x\in B$, then this implies $z=(x,y)\in B\times C$, contrary to $z\notin B\times C$. Hence $x\notin B$, i.e. $x\in A-B$ and together with $y\in C$, we have $z=(x,y)\in(A-B)\times C$. Since $z$ was arbitrary, we conclude $(A\times C)-(B\times C)\subseteq (A-B)\times C$.

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@Mathematics: And you have $A\times (B-C)= (A\times B)-(A\times C)$ as well. ;-) –  B. S. Oct 20 '12 at 18:12
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