Translation to language P (Predicate Logic)

Question

I am taking an intro class in philosophy and I have having trouble with some assignment questions. I need to translate into the language P.

Here's the translation keys:

Fx: x is a firefighter Dx: x is a doctor Nx: x is an nurse Gx: x plays guitar Lxy: x likes y a: Annie b: Bob

I am managed to answer other ones, but I am having trouble with these 8. Explanation will be greatly appreciated!

q8. All firefighters like those doctors who play guitar.

q21. Some doctors don’t like any firefighter who is a nurse.

q29. Bob doesn’t like any doctor who is either a nurse or plays guitar.

q36. Bob likes all those doctors who like Annie.

q39. Some firefighters who like Bob don’t like Annie.

q44. All nurses who like Annie but don’t like Bob play guitar.

q49. Every firefighter likes some doctors who don’t like Annie.

q50. No doctor who plays guitar like all firefighters.

Similar for 44 : $\forall x [(Nx \land Lxa \land \lnot Lxb) \to Gx)]$ — Mauro ALLEGRANZA, Nov 18 '19 at 15:20
For 8 we have : $\forall x \ \forall y \ [(Fx \land Dy \land Gy) \to Lxy]$. — Mauro ALLEGRANZA, Nov 18 '19 at 15:37
$\exists x \ \forall y \ [(Dx \land Dy \land Ny) \to \lnot Lxy]$. — Mauro ALLEGRANZA, Nov 18 '19 at 15:46
The "ingredients" are always the same; if you have a binary predicate, like $Lxy$, in principle you need two different quantifiers. They must be chosen according to the English sentence : $\forall$ for "Every" and $\exists$ for "Some", playing with the negation sign to manage the various cases : $\lnot \exists$ is "No/none". — Mauro ALLEGRANZA, Nov 18 '19 at 15:56
@MauroALLEGRANZA I am able to answer all the other one, beside this last one :q50. No doctor who plays guitar like all firefighters. hint? Thank you — GarlicSTAT, Nov 18 '19 at 19:22
I'm voting to close this question as off-topic because it is already answered and accepted in PhSE — Mauro ALLEGRANZA, Nov 19 '19 at 10:10
In general asking multiple-part problems with no indication of work that relates the various parts will suggest to your Readers that you are merely passing through an assigned exercise with little or no forethought. If there is some specific difficulty you encountered with those parts, it will typically first manifest in one part of the exercise, and articulating that difficulty is important for Readers willing to help you overcome it (aka learning the math behind the exercise) rather than providing a full set of "cut and paste" solutions). — hardmath, Nov 19 '19 at 19:22

score 2 · Accepted Answer · edited Jun 12 '20 at 10:38

With these translations, it is often helpful to keep in mind the folowing 4 Aristotelan sentences and their predicate logic translations

A: "All $S$ are $P$": $\forall x (S(x) \to P(x))$

I: "Some $S$ are $P$": $\exists x (S(x) \land P(x))$

E: "No $S$ are $P$" (which is the same as saying that "All $S$ are not $P$"): $\forall x (S(x) \to \neg P(x))$

O: "Some $S$ are not $P$": $\exists x (S(x) \land \neg P(x))$

Why are these helpful? It is because many English sentences are of one of these forms. That is, many English sentences have a subject term (that which the claim is about), and a predicate term (that which you say about those things). This is why I used $S$ and $P$ in the statements above: (S)ubject term and (P)redicate term. The letters are just a handy way to quickly refer to these 4 different forms.

So, for example, let's consider:

q49. Every firefighter likes some doctors who don’t like Annie.

Now, what the subject term of this sentence? Who or what are we making a claim about? It is firefighters. OK, and what do we say about firefighters? That they like some doctors who don’t like Annie. OK, so the subhject term is easy, but the predicate term is still complex. Well, don't panic! For now, just break it down into subject term and predicate term. Indeed, it is this 'divide and conquer' method that will eventually get you there.

OK, now, what is the basic form? Well, it says that something is true for all firefighters, and so it is of the A form, and so we get:

$\forall x (S(x) \to P(x))$

where

$S(x)$: $X$ is a firefighter

$P(x)$: $x$ likes some doctors who don't like Annie

Now, again, the subject term is easy, that's just $F(x)$

But the predicate term is more complex. So, how do we translate "$x$ likes some doctors who don't like Annie"? Well, it's a little less obvious, but also this sentence is one of the Aristotelean forms: notice that while this claim seems is about $x$, we can also treat it as a claim about doctors who don;t like Annie. That is, we can paraphrase this claim as:

"Some doctors who don't like Annie are liked by $x$"

And that we recognize as an I claim:

$\exists x (S(x) \land P(x))$

where:

$S(x)$: $x$ is a doctor who does not like Annie

$P(x)$: $x$ is not liked by .... $x$....?

OK, here's a little issue: since we already use $x$ for a firefighter who is going to like a doctor, we should use a different variable for the doctor. OK, so let's use:

$\exists y (S(y) \land P(y))$

where:

$S(y)$: $y$ is a doctor who does not like Annie

$P(y)$: $y$ is not liked by $x$ ... Ah! Now it works nicely!

OK, so let's translate these now:

$S(y)$: $D(y) \land \neg L(y,a)$

$P(y)$: $L(x,y)$

OK, great! We got all the parts .. now we just plug it all into the already identified forms. That is:

"Some doctors who don't like Annie are liked by $x$"

translates as:

$\exists y ((D(y) \land \neg L(y,a)) \land L(x,y)$

And, plugging that into the form:

$\forall x ("$x$\text{ is a firefighter}" \to "$x$\text{ likes some doctors who don't like Annie}")$

which we paraphrased as:

$\forall x ("$x$\text{ is a firefighter}" \to \text{ "Some doctors who don't like Annie are liked by} $x$")$

we get:

$\forall x (F(x) \to (\exists y ((D(y) \land \neg L(y,a)) \land L(x,y)))$

Translation to language P (Predicate Logic)

1 Answers1