Simple Cartesian Product Proof(s)

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SUMMARY

The discussion centers on proving the equality of the Cartesian product Ax(B^C) and the intersection (AxB)^(AxC) for any sets A, B, and C. The proof demonstrates that an ordered pair (x,y) belongs to Ax(B^C) if and only if x is an element of A and y is an element of both B and C. The logical steps confirm that the proof is valid, and the concern about redundancy in stating "x is an element of A" is addressed, affirming that it does not constitute poor form in mathematical proof.

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Homework Statement



Prove for all sets A, B, and C that:
Ax(B^C) = (AxB)^(AxC)


Homework Equations


AxB is the Cartesian Product of A and B. That is, the set of all ordered pairs (x,y) such that x is an element of A and y is an element of B.
^ denotes intersection (and)

The Attempt at a Solution



(x,y) is an element of Ax(B^C)
iff x is an element of A and y is an element of (B^C)
iff x is an element of A and y is an element of B and y is an element of C
iff (x is an element of A and y is an element of B) and (x is an element of A and y is an element of C)
iff (x,y) is an element of (AxB) and (x,y) is an element of (AxC)
iff (x,y) is an element of (AxB)^(AxC)

I'm confident that I'm "right," but I did sort of produce a second "x is an element of A" out of nowhere to tidy up my sequence of logic. Is this poor form? What would be better?
 
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This is alright, and not at all "poor form". If x \in A, then also x \in A \wedge x \in A.
 

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