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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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  • $\begingroup$ I typeset your question to make it easier to read - please check I didn't make any mistakes in the notation. $\endgroup$
    – mdp
    Commented Oct 8, 2012 at 16:56
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    $\begingroup$ "The answer" is incorrect -- for example it is true for the interpretation where $P$ hold s for all three individuals, simply because then $P(a)$ is true and therefore $P(a)\lor\mathit{whatever}$ is automatically true too. $\endgroup$ Commented Oct 8, 2012 at 17:04
  • $\begingroup$ yes, that correct thank for the correction $\endgroup$
    – Samuel
    Commented Oct 9, 2012 at 4:26

1 Answer 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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  • $\begingroup$ But suppose it is true that $a = b$ ... then the l.h.s. could be true and the r.h.s. false. $\endgroup$ Commented Oct 8, 2012 at 18:07
  • $\begingroup$ @Peter: Clearly we are to assume that the objects $a,b$, and $c$ are distinct. $\endgroup$ Commented Oct 8, 2012 at 18:10
  • $\begingroup$ @BrianScott Indeed we are! So the question as set is underspecified. $\endgroup$ Commented Oct 8, 2012 at 18:55
  • $\begingroup$ Thank Brian very clear explaintion $\endgroup$
    – Samuel
    Commented Oct 9, 2012 at 4:33
  • $\begingroup$ @user1660416: You’re very welcome. $\endgroup$ Commented Oct 9, 2012 at 6:30

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