Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

May I ask to have a look at the following proof I did. Thanks.

Question:

$(A \cup B)\cap ( A\cup C)=A \cup (B \cap C)$

My Attempt:

A$\cup$(B$\cap$C):

$\forall$x[(x$\in$A)]$\vee$[(x$\in$B)(x$\in$C)]

=$\forall$x[(x$\in$A)$\vee$(x$\in$B)]$\wedge$[(x$\in$A)$\vee$(x$\in$C)]

=$\forall$x[(x$\in$A)$\cup$(x$\in$B)]$\cap$[(x$\in$A)$\cup$(x$\in$C)]

=(A$\cup$B)$\cap$(A$\cup$C)

share|improve this question
    
i see no problem –  Aang Mar 6 '13 at 16:45
    
@Avatar: Maybe you would like to inform amWhy to make his answer a comment. –  azimut Mar 6 '13 at 17:04
    
i would have, but people don't seem to listen to me. –  Aang Mar 6 '13 at 17:07
    
@azimut: But you see, he had something to add. –  Aang Mar 6 '13 at 17:07

2 Answers 2

up vote 5 down vote accepted

Very well done. The "gist" of your logic is correct. There are only a few minor problems.

One thing I'd add, and it may simply have been a typo, is in the first line of your proof, you want to add a missing $\land$ between $(x\in B)(x\in C)$:

$$x \in [A \cup(B\cap C)] \iff [(x\in A) \lor ( x\in B \land x \in C)]\tag{*}$$

Notice I also used $x \in [A\cup(B\cap C)]$ to start, with no need for the universal quantifier, because we are making claims about precisely any/every element belonging the set in question. We use this notation since we are aiming to show

$$x \in A\cup(B\cap C) \iff x\in [(A\cup B) \cap (A \cup C)]$$ and in doing so, we will have proven the desired equality: $$A \cup(B\cap C) = [(A\cup B) \cap (A \cup C)]$$

So you want to end with $x \in [(A\cup B) \cap (A \cup C)]$ to finish the proof of the equality of the Left hand side and the right hand side.

And use $\iff$ between lines (as used in (*)). That means that "if and only if".

share|improve this answer

The idea of your proof is correct. But it's not written down in a correct way.

Look at the first equality sign. You write $$A \cup (B\cap C) = \forall x[(x\in A)]\vee [(x\in B)(x\in C)].$$ On the left hand side of the equality, there is a set. But on the right hand side there is not a set, but a logical expression. This is certainly not ok.

What you thought of is probably $$A \cup (B\cap C) = \{x \mid x\in A\vee (x\in B \wedge x\in C)\}.$$ Now on both sides of the equation, there is a set. And by definition of $\cap$ and $\cup$, they are the same.

I hope you get the idea. Try to rewrite your proof accordingly.

share|improve this answer
    
+1 Thanks azimut –  Software Mar 6 '13 at 17:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.