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I cannot understand the following solution from my tutorial note.

The question is like this:

Let $P(x)$ be a predicate with universe of discourse $\{a,b,c\}$. The quantifier $\exists!$ is used to assert that there is a unique element of the universe of discourse which makes a predicate true. Now express $\exists ! xP(x)$ using only the operators $\land$, $\lor$ and $\lnot$.

The answer is:

$\exists ! xP(x) \equiv P(a) \lor P(b) \lor P(c) \lor [\lnot P(a) \land P(b) \land \lnot P(c)] \lor [P(a) \land \lnot P(b) \land \lnot P(c)]\lor [\lnot P(a) \land \lnot P(b) \land P(c)]$

What is the meaning of the last three parts?

$\lor [\lnot P(a) \land P(b) \land \lnot P(c)] \lor [P(a) \land \lnot P(b) \land \lnot P(c)]\lor [\lnot P(a) \land \lnot P(b) \land P(c)]$

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    yes, that correct tha$n$k for the correctio$n$2012-10-09

1 Answers 1

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The answer is wrong. It should read:

$\exists!xP(x)\equiv [\lnot P(a) \land P(b) \land \lnot P(c)] \lor [P(a) \land \lnot P(b) \land \lnot P(c)]\lor [\lnot P(a) \land \lnot P(b) \land P(c)]$

The first term in square brackets says that $b$ has the property $P$ and $a$ and $c$ do not; the second says that $a$ has the property and $b$ and $c$ do not; and the third says that $c$ has the property and $a$ and $b$ do not. Since $a,b$, and $c$ are the only objects in the universe of discourse, these are the only three ways that exactly one of them can have the property $P$.

It should not include $P(a)\lor P(b)\lor P(c)$: that part alone is equivalent to $\exists xP(x)$, and $\exists xP(x)\lor\exists!xP(x)$ is equivalent to $\exists xP(x)$, not to $\exists!xP(x)$.

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    @user1660416: You’re very welcome.2012-10-09