(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

1) [tex] P\rightarrow Q \equiv \neg P \vee Q[/tex]

2) [tex] \neg(P\vee Q)\equiv \neg P \wedge \neg Q [/tex]

3) [tex] P \vee (Q\vee R) \equiv (P\vee Q) \vee R \equiv P \vee Q\vee R [/tex]

3. The attempt at a solution

[tex]\exists xAP(x)\vee\exists xBP(x)[/tex]

[tex]\exists x(x\in A \rightarrow P(x))\vee\exists x(x\in B \rightarrow P(x))[/tex] definition of the above

[tex]\exists x[(x\notin A \vee P(x))\vee (x\notin B \vee P(x))][/tex] using (1)

[tex]\exists x(x\notin A \vee x\notin B \vee P(x))[/tex] using (3)

[tex]\exists x[\neg(x\in A \wedge x\in B) \vee P(x)][/tex] using (2)

[tex]\exists x[x\in (A\cap B) \rightarrow P(x)] [/tex] using (1) and simplifying the expression ... [tex]\exists x(A\cap B)P(x)[/tex]

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:

[tex]\exists xAP(x)\wedge\exists xBP(x)[/tex] 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

[tex]\exists xAP(x)\vee\exists xBP(x)[/tex] 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 [tex]\exists x(A\cap B)P(x)[/tex] 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: [tex]\exists x(A\cup B)P(x)[/tex] if there exists a certain shape OR a certain colour then P likes the object

Can anyone show me where I'm going wrong?

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# Homework Help: Quantifier equivalence in set theory

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