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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.

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    No attempts done ? – Mauro ALLEGRANZA Nov 18 '19 at 15:09
  • Try with the first one : "All firefighters..." must be "For every x (if Fx ...)" – Mauro ALLEGRANZA Nov 18 '19 at 15:10
  • The next one is similar : "Some doctors ..." will be "Exists x such that (Dx and ...)" – Mauro ALLEGRANZA Nov 18 '19 at 15:12
  • there are 50 of theses question, I am able to do the other ones, just these 8 questions I have no idea. – GarlicSTAT Nov 18 '19 at 15:12
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    Maybe it is useful to specify that "language P" is predicate logic language. – Mauro ALLEGRANZA Nov 18 '19 at 15:13
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Many predicate language sentences come in one of two forms: "All A's are B's" and "Some A's are B's". There's also "No A's are B's", but that's just a negation of "Some A's are B's". They are translated as:

All A's are B's: ∀x(Ax -> Bx)

and

Some A's are B's: ∃x(Ax & Bx)

All of your sentences in one way or another have one of those two basic forms. Take q8 for example:

q8. All firefighters like those doctors who play guitar.

This has the "All A's are B's" form. We have A = "firefighters" and B = "like those doctors who play guitar". So the translation should look like this:

∀x(Fx -> (x likes those doctors who play guitar))

Now we have to translate that second part. This part also fits into one of the basic forms: "x likes those doctors who play guitar" is the same as "All doctors who play guitar are liked by x", which is of the form "All A's are B's". In this case A = "doctors who play guitar" and B = "liked by x" So it should be:

∀y((Dy & Gy) -> Lxy)

Putting the two parts together we get:

∀x(Fx -> ∀y((Dy & Gy) -> Lxy))

Let's do another one:

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

This has the "Some A's are B's" form, where A = "doctors" and B = "don't like any firefighter who is a nurse". So we get:

∃x(Dx & (x doesn't like any firefighter who is a nurse))

Next we should translate the second part. We should try to see if it fits one of the two basic forms. It does: "x doesn't like any firefighter who is a nurse" is the same as "No firefighter is liked by x" which has the form ~∃x(Ax & Bx), where A = "firefighter and nurse" and B = "liked by x". So for this we get:

~∃y((Fy & Ny) & Lxy)

Combining the two parts we get:

∃x(Dx & ~∃y((Fy & Ny) & Lxy))

I hope you can generalize from this to the other questions.

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