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How do I express this sentence in predicate logic, when I can pick the domain of discourse myself?

Friends of Michelle's friends are her friends.

I was thinking of picking the domain of discourse Michelle's friends:

So that I would get $\forall x (Fx \rightarrow x) $

$F =$ friends of

Is this correct?

How would I express this?

Thanks in advance, Rope.

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  • $\begingroup$ @Rope As I see it you got two sensible options. One is to consider the domain of discourse as all people, the other is to consider the domain of discourse as being Michelle's friends. Both of these with the implicit assumption that Michelle is only friend with people and not animals, playstations or whatever. In any of those cases, you need to attribute to Michelle a name in the language. Usually constants are represented by small letters from the beginning of the alphabet, but it is legitimate to pick $m$ for Michelle. Now just formalize it. $\endgroup$ – Git Gud Jun 2 '14 at 14:10
  • $\begingroup$ Note that $\forall x(Fx\to \color{blue}x)$ isn't even a well-formed formula. For it to be a well formed formula, $\color{blue}x$ needs to be a constant or a predicate. It is neither. $\endgroup$ – Git Gud Jun 2 '14 at 14:11
  • $\begingroup$ Hmm.. I thought that the x wasn't correct. But I don't know how to pick an appropriate domain nor how to show the friends of friends relation in a way that makes sense. $\endgroup$ – Byebye Jun 2 '14 at 14:20
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    $\begingroup$ If the domain is all people, you can formalize it as $\forall x\forall y((Fxm\land Fyx)\to Fym)$, where $Fxy$ is the binary predicate $\text{$x$ is a friend of $y$}$. In loglish (english fused with logic - I've seen Peter Smith use this term): given two people $x$ and $y$, if $x$ is a friend of Michelle and $y$ is a friend of $x$, then $y$ is a friend of Michelle. $\endgroup$ – Git Gud Jun 2 '14 at 14:25
  • $\begingroup$ Aaaaah of course! Thanks! (Sorry I didn't read your comment before, I deleted the other comment as it made no sense :).) $\endgroup$ – Byebye Jun 2 '14 at 14:31
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After the back and forth with GitGud in the comments I got the following answer.

You're allowed to put forth multiple for-all variables (which I didn't consider before).

The domain of discourse has to be all People or all humans.

It follows that for the Friends of relation we can put $Fxy$ as x is a Friend of y. Hence we get:

$$\forall x \forall y((Fxm \wedge Fyx)\rightarrow Fym)$$

$Fxm$ All humans that are not y are friends of michelle.

$Fyx$ all humans that are not x are friends of friends of michelle(stated in the previous predicate)

$\rightarrow$ thus it is the case

$Fym$ that all friends of friends of michelle are friends of michelle

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