I used the definitions, now what?

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Homework Help Overview

The discussion revolves around proving set identities involving intersections and unions in set theory. The original poster presents two statements to prove, focusing on the logical equivalences of set membership.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to manipulate the definitions of set operations to show equivalence but expresses uncertainty about a logical leap needed for the first statement. They also question what logical conclusion is missing for the second statement.

Discussion Status

Some participants suggest using a truth table to analyze the logical relationships, indicating that there may be multiple ways to approach the proofs. One participant provides a reasoning path for the second statement, suggesting that the sets are equal by demonstrating mutual inclusion. The original poster acknowledges the guidance received and indicates progress.

Contextual Notes

The discussion does not provide complete information on the definitions or properties of the sets involved, nor does it clarify any specific homework constraints that may apply.

JinM
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Homework Statement



Prove:

1. [tex]R \cap ( S \cup T ) = (R \cap S) \cup (R \cap T)[/tex]

2. [tex]S \cap ( S \cup T ) = S[/tex]

2. The attempt at a solution

I suppose this is all about using the definitions, and I eventually get down to this:

For (1), the LHS is down to x e R and (x e S or x e T), while the RHS is (x e R and x e S) or (x e R and x e T). There's one small leap here, I know. How do I show these two are equivalent?

For (2), I should show that (x e S) and (x e S or x e T) is equivalent to (x e S). What logical conclusion am I missing here?
 
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If nothing else, use a truth table. x is either in or is not in R, S and T. That leaves you eight cases. Four in the second one. Is the logic true in all cases? With some moderate cleverness you don't even have to check all eight. Sure, it's just logic.
 
for (2), if x is in the lhs then x is in S and (S or T). So x is in the rhs.
If x is in the rhs, x is in S and so x is in (S or T). So x is in S and (S or T), so x is in the lhs.

Thus the lhs is a subset of the rhs and the rhs a subset of the lhs, so the two sets are equal.
 
Thanks guys -- that got me on the right track.
 

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