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Statement:

Jane saw a police officer and Roger saw one too

Let

  • $J(x)$: Jane saw police officer $x$

  • $R(x)$: Roger saw police officer $x$

Hence,

$$\exists x \exists y \left( J(x) \land R(y) \right)$$

Is this correct?

Rodrigo de Azevedo
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Jessica Griffin
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    You could simplify it to $P(z)$ is true if "$z$ saw a police officer", and then have $\exists r, j: P(r) \wedge P(j)$. –  Jun 28 '20 at 15:55
  • And even more - you can consider sentence "x saw y" and use 4 variables. – zkutch Jun 28 '20 at 15:59
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    Your answer is not elegant. You need to state $\exists x P(x)$, where $P(x)$ stands for "$x$ is a police officer." This is from Chapter 2 of Velleman's How to Prove it and the solution is given at the end of the book. – toronto hrb Jun 28 '20 at 16:10
  • @torontohrb i have seen the solution but i want to know if this is correct too – Jessica Griffin Jun 28 '20 at 16:17
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    @JessicaGriffin This is correct. – Anurag A Jun 28 '20 at 16:20
  • @zkutch $4$ variables or a $2$-tuple taking values in a set of cardinality $4$? – Rodrigo de Azevedo Jun 28 '20 at 16:21
  • Nice suggestion. – zkutch Jun 28 '20 at 16:22
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    @RodrigodeAzevedo Notice that this may not need to have a model with 4 elements. It can also have a model o size 1. Just consider the case Roger = Jane and is a police officer. Also If Roger saw Jane and Jane did the same and both are officers we can have a model of size two, the case of three elements could be where we have jane roger and just one police officer, therefore jane and roger saw the same officer. – edgar alonso Jun 28 '20 at 16:44
  • @edgar alonso But that's not a problem. A structure in which Jane and Roger are the same person, who also happens to be a police officer, who saw themself, is perfectly consistent with the original sentence and should be permitted by the translation. – Natalie Clarius Jun 29 '20 at 22:58

3 Answers3

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Your translation is not incorrect, but quite imprecise. What you summarized as one predicate, "$P(x)$ = Jane saw police officer ", still contains an individual (Jane), a relation (saw), and a property (is a police officer). But it is not good practice to import specific individuals (Jane) and additional restrictions (is a police officer) into a predication (z saw y). To exaggerate, you could of course just shove everything into one symbol and define a single zero-place predicate "$P()$ = Jane saw a police officer and Roger saw one too" and be done with it -- technically not wrong, but kinda defeats the point of it all. Atomic formulas should express atomic propositions; what can be decomposed in more detail should be decomposed in more detail.

So analyze the sentene further into the parts mentioned above:

  • individuals:
    • $j$ = Jane
    • $r$ = Roger
  • properties and relations:
    • $po(x)$ = $x$ is a police officer
    • $saw(x,y)$ = $x$ saw $y$

As a rule of thumb,

$$\text{an A (is) B/there is an A that B}$$

translates as

$$\exists x (A(x) \land B(x))$$

and

$$\text{all A (are) B/everything that A B}$$

translates as

$$\forall x (A(x) \to B(x))$$

What you want here is "$\text{ a } \underbrace{\text{police officer}}_{A} \text{ is } \underbrace{\text{seen by Jane}}_{B}$", and the same with Roger. So you end up with

$$\exists x (po(x) \land saw(j,x)) \land \exists y (po(y) \land saw(r,y))$$

If you prefer, you can write the formula in a logically equivalent variant with the quantifiers moved to the front and the conjuncts commuted:

$$\exists x \exists y (po(x) \land po(y) \land saw(j,x) \land saw(r,y))$$

What is important is, as you did, to introduce two different individuals for the police officers and apply the predications to both of them, because the original sentence doesn't say that Jane and Roger necessarily saw the same officer (though the above formula correctly doesn't exclude this possibility, either).

Natalie Clarius
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If the question was "Jane saw every police officer", would $\lnot \exists x . \lnot P(x)$ be correct answer? That is saying "there is not a police officer that Jane didn't see" which while technically equivalent, isn't actually the hypothetical question.

$$\exists x \exists y \left( J(x) \land R(y) \right)$$

This is technically correct, but it is as if you were translating "there were some police officers seen by Jane and Roger." The statement is "Jane saw a police officer and Roger saw one too", so you would expect the answer to be something like $\text{this } \land \text{ that}$. I would suggest:

$$(\exists x. J(x)) \land (\exists x.R(x))$$

DanielV
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  • I don't see how the Asker is saying "there is not a police officer that Jane didn't see". As you say, shouldn't that be $\neg \exists x \neg J(x)$ or $\forall x J(x)$. – Jam Jun 29 '20 at 17:12
  • @Jam IF that was the question. The point is that even though something may be logically equivalent, it isn't necessarily a good translation. – DanielV Jun 29 '20 at 17:32
  • I see. That should probably be "it is as if you were saying" rather than "you are saying" then. – Jam Jun 29 '20 at 17:36
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I wouldn't call your solution incorrect per se but it's not an ideal answer. The functions should be predicates, which are essentially clear yes/no questions. The desired predicate (named $P$ in the comments) asks "did $x$ see a police officer", to which "yes" implies $x$ did, while "no" implies $x$ didn't. But your predicate, $J$, asks "did Jane see that police officer" - it's not as informative what yes or no mean here. "No" could mean Jane didn't see the police officer (but someone else did) or it could mean Jane didn't see that police officer (but she did see a different one). This is why $P$ is better.

Jam
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  • If the downvoter could care to explain what they dislike about my answer, I would appreciate it. – Jam Jun 28 '20 at 17:13
  • If you are trying to make the claim that a different set of predicates more naturally translates the problem, please say so plainly. – DanielV Jun 29 '20 at 16:51
  • @DanielV I believe I did say so plainly: I said what was unnatural about their predicate. – Jam Jun 29 '20 at 17:12
  • Where do you see function symbols in the solution suggested by the OP? And why would $P$ (a zero-place predicate?) be more informative than $P(x)$? – Natalie Clarius Jun 29 '20 at 22:13
  • @lemontree I don't understand your point. A predicate is a function - by this I was referring to the Asker's $J(x)$ and $R(x)$. Also, $P$ was referring to the predicate itself as opposed to its output, a la notation for functions. I did not intend to suggest that $P(x)$ was a zero-place predicate but I do not believe I did so. If my notation was irregular then let me know. – Jam Jun 29 '20 at 22:27
  • It is confusing to talk about functions in the context of predicate logic, because one would think that you mean function symbols, which are a different thing from predicate symbols. That boolean (yes/no) functions are to be expressed as predicates does not need explicit mention, this is how predicate logic works and what the OP already did. So I don't understand what you're suggesting to change, and why this would be better than OP's attempt. – Natalie Clarius Jun 29 '20 at 22:53