Quantifier equivalence in set theory

[GregA]

Homework Statement

I have been asked to show that \exists xAP(x)\vee\exists xBP(x) is equivalent to \exists x(A\cup B)P(x)

Homework Equations

1) P\rightarrow Q \equiv \neg P \vee Q
2) \neg(P\vee Q)\equiv \neg P \wedge \neg Q
3) P \vee (Q\vee R) \equiv (P\vee Q) \vee R \equiv P \vee Q\vee R

The Attempt at a Solution

\exists xAP(x)\vee\exists xBP(x)
\exists x(x\in A \rightarrow P(x))\vee\exists x(x\in B \rightarrow P(x)) definition of the above
\exists x[(x\notin A \vee P(x))\vee (x\notin B \vee P(x))] using (1)
\exists x(x\notin A \vee x\notin B \vee P(x)) using (3)
\exists x[\neg(x\in A \wedge x\in B) \vee P(x)] using (2)
\exists x[x\in (A\cap B) \rightarrow P(x)] using (1) and simplifying the expression ... \exists x(A\cap B)P(x)

This is not what the book is asking me to show though!...and I can't see where I've gone wrong either ...Starting from the RHS and trying to show it is equivalent to the LHS gets me:
\exists xAP(x)\wedge\exists xBP(x) which is still no good!

Trying to think of it in terms of words then:
Referring to A as (the set of colours of an object), referring to B as (the set of shapes of an object) and P(x) as ( P likes it) then
\exists xAP(x)\vee\exists xBP(x) is saying that if there exists a certain colour then P likes the object or if there exists a certain shape then P likes the object
My conclusion \exists x(A\cap B)P(x) however says that if there exists a certain shape AND a certain colour then P likes the object

what I am meant to show however seems correct inspite of my efforts ie: \exists x(A\cup B)P(x) if there exists a certain shape OR a certain colour then P likes the object

Can anyone show me where I'm going wrong?

 

[HallsofIvy]

Homework Statement

I have been asked to show that \exists xAP(x)\vee\exists xBP(x) is equivalent to \exists x(A\cup B)P(x)

That is: "either there exist x in A such that P(x) is true or there exist x in B such that P(x) is true" is equivalent to "there exist x in A union B such that P(x) is true"

Homework Equations

1) P\rightarrow Q \equiv \neg P \vee Q
2) \neg(P\vee Q)\equiv \neg P \wedge \neg Q
3) P \vee (Q\vee R) \equiv (P\vee Q) \vee R \equiv P \vee Q\vee R

The Attempt at a Solution

\exists xAP(x)\vee\exists xBP(x)

Yes, that is one of the two statements

\exists x(x\in A \rightarrow P(x))\vee\exists x(x\in B \rightarrow P(x)) definition of the above

Are you sure that's the same thing? I would interpret this as "either there exist x such that if x is in A then P(x) is true or there exist x such that if x is in B then P(x) is true. If P(x) were false for all x in A or B, both statements would be true!

Since x in A \rightarrow x in A\cup B and x in B \right arrow x in A\cup B,
(\exists xAP(x)\vee\exists xBP(x))\rightarrow x in A\cup B P(x)

 

[courtrigrad]

\exists x(x\in A \rightarrow P(x))\vee\exists x(x\in B \rightarrow P(x))should be \exists x(x\in A \vee x\in B) \rightarrow P(x))

 

[GregA]

Are you sure that's the same thing? I would interpret this as "either there exist x such that if x is in A then P(x) is true or there exist x such that if x is in B then P(x) is true. If P(x) were false for all x in A or B, both statements would be true!

I see your point!...thanks for that
Will now return to my book, figure out where I mis-understood it and then try to finish this question properly

*edit* and thankyou to you Courtigrad as well...I was totally unaware of where I was messing up