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Kolmin
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Homework Statement
Prove that
[itex]A[/itex][itex]\times[/itex][itex]([/itex][itex]B[/itex][itex]\cap[/itex][itex]C[/itex][itex])[/itex][itex]=[/itex][itex]([/itex][itex]A[/itex][itex]\times[/itex][itex]B[/itex])[itex]\cap[/itex][itex]([/itex][itex]A[/itex][itex]\times[/itex][itex]C[/itex][itex])[/itex]
In particular what I cannot prove is
[itex]([/itex][itex]A[/itex][itex]\times[/itex][itex]B[/itex])[itex]\cap[/itex][itex]([/itex][itex]A[/itex][itex]\times[/itex][itex]C[/itex][itex])[/itex][itex]\subseteq[/itex][itex]A[/itex][itex]\times[/itex][itex]([/itex][itex]B[/itex][itex]\cap[/itex][itex]C[/itex][itex])[/itex]
Homework Equations
The Attempt at a Solution
In order to learn how to deal with proofs I am reading How to prove it: a structured approach, and I am using Proof Designer. This is the main problem I think...
First of all, I tried all possible approaches, but still I cannot figure out how to deal the problem I have, which I will explain.
I found various direct proofs of this biconditional (even on the book itself) and, given
[itex]\exists[/itex][itex]a[/itex][itex]\in[/itex][itex]A[/itex][itex]\exists[/itex][itex]b[/itex][itex]([/itex][itex]b[/itex][itex]\in[/itex][itex]B[/itex][itex]\wedge[/itex][itex]p[/itex][itex]=[/itex][itex]([/itex][itex]a[/itex][itex],[/itex][itex]b[/itex][itex])[/itex][itex])[/itex]
[itex]\exists[/itex][itex]a[/itex][itex]\in[/itex][itex]A[/itex][itex]\exists[/itex][itex]b[/itex][itex]([/itex][itex]b[/itex][itex]\in[/itex][itex]C[/itex][itex]\wedge[/itex][itex]p[/itex][itex]=[/itex][itex]([/itex][itex]a[/itex][itex],[/itex][itex]b[/itex][itex])[/itex][itex])[/itex]
all assume that this [itex]b[/itex] has to be the same for [itex]B[/itex] and [itex]C[/itex].
Indeed, if this is the case, the proof is quite straightforward, BUT...we cannot assume so - at least using Proof Designer. As a matter of fact, the software, using the existential istantiation, asks you to choose a certain name (let's say [itex]a[/itex][itex]_{0}[/itex]) and that's the problem. In the first expression I can put [itex]a[/itex][itex]_{0}[/itex] and [itex]b[/itex][itex]_{0}[/itex], but in the second one, I cannot force the program to use the same variables, because they are already being used, so I have to use something like [itex]a[/itex][itex]_{1}[/itex] and [itex]b[/itex][itex]_{1}[/itex]. And that's the problem I guess, cause if I could put [itex]a[/itex][itex]_{0}[/itex] and [itex]b[/itex][itex]_{0}[/itex], the proof would be trivial.
Quite instructive, in order not to use the same object for two different things simply because it helps us to get a wrong proof, but I cannot figure out how to get the right proof!