MHB How Can I Prove the Set Operation AX(BΔC) = (AXB)Δ(AXC)?

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The discussion focuses on proving the set operation AX(BΔC) = (AXB)Δ(AXC). The proof begins by considering an element (x,y) in A×(BΔC) and analyzing the implications of y being in either B or C, but not both. It demonstrates that if y is in B and not in C, then (x,y) belongs to A×B but not to A×C, thus confirming (x,y) is in (A×B)Δ(A×C). A similar argument is made for the case where y is in C but not in B. The conclusion invites further exploration of proving the converse inclusion.
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Hello
There is the possibility that they help me to solve this demonstration. please

AX(BΔC)=(AXB)Δ(AXC)
 
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Hi, and welcome to the forum.

daniel felipe said:
AX(BΔC)=(AXB)Δ(AXC)
Suppose that $(x,y)\in A\times(B\triangle C)$. Then $x\in A$ and $y\in B\triangle C$. The latter means that $y\in B$ or $y\in C$, but not both. In the first case, i.e., $y\in B$ but $y\notin C$, we have $(x,y)\in A\times B$. However, $(x,y)\notin A\times C$ because that would mean, in particular, that $y\in C$. Therefore, $(x,y)\in (A\times B)\triangle (A\times C)$. The second case ($y\in C$ but $y\notin B$) is considered similarly. This concludes the proof that $A\times(B\triangle C)\subseteq (A\times B)\triangle (A\times C)$. You can try proving the converse inclusion.

For the future, please read the http://mathhelpboards.com/rules/, especially rule 11 (click "Expand" button on top).
 
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