Philosophy 2340
Symbolic Logic

Advice on Chapter 9 Problems

Note that most of my hints begin with a translation prefixed with ">". These are incorrect solutions. The rest of the hint explains why they are wrong and makes positive suggestions about how to fix them.

9.2.6

> Ax Ay ((Cube(x) & Cube(y)) -> Ex Ey~Larger(x, y))

Notice that your sentence means "if everything is a cube, then there are objects x and y such that x is not larger than y" -- probably not what you intended!


9.3.6

> (Cube(a) & Small(a))

Bozo was attempting to say that there is a small cube, but not that any particular cube is small. His mistake is using a constant where he needs a variable.


9.3.10

> Ax (Tet(x) & Small(x)) -> Ax FrontOf(x, e) 

Notice that in English, your sentence says "if everything is a small tetrahedron, then everything is in front of e" -- probably not what you intended!


9.12.3

> Ex (Dodec(x) -> (Small(x) v Medium(x) v Large(x)))

remember, an existentially quantified conditional will virtually never be the right translation for an English sentence. If you see you've written something of the form Ex (S(x) -> P(x)), it's time to go back and rethink!

In this case, the sentence is about *every* dodecahedron, so you want a universal quantifier rather than an existential.

> Ax (Dodec(x) & Large(x)) v Ax (Dodec(x) & Medium(x)) v Ax (Dodec(x) & Small(x))

This is pretty close! You're supposed to say that all dodecahedra are either large, medium, or small. Your sentence actually says that *everything* is a large, medium, or small dodecahedron! But that's too strong. What you need is: Ax(Dodec(x) ->( Large(x) v Medium(x) v Small(x))).


9.12.8

> Ex (Dodec(x) & ~Large(x)) & Ex (Dodec(x) & ~Small(x))

Again, pretty close. This says that there's a dodec that's not large and there's a dodec that's not small. But it doesn't imply that they are the same dodec. You need: Ex (Dodec(x) & ~(Large(x) v Small(x))) or something equivalent to this.


9.16.13

> ~Ex (Dodec(x) & Large(x))

Your sentence says that there isn't something that is a large dodec (= there are no large dodecs); it needs to say that there *is* something that *isn't* a large dodec. (Notice that, while your sentence will be false if anything is a large dodec, the sentence to be translated can be true even if there is a large dodec, provided that there is something else that isn't a large dodec.)

> Ex (Dodec(x) & ~Large(x))

The English says "something is not a large dodec." This would be true in a world containing only one object, a small cube. But your sentence would be false in that world. See why?


9.16.14

> Ex ~(Large(x) & Dodec(x))

The English sentence was "it's not the case that something is a large dodecahedron." Notice that this would be true, for example, in a world with *nothing* in it. However, your translation says "something is not a large dodecahedron." Notice that your translation would *not* be true in a world with nothing in it.

Also, your translation would be true in a world that contained a large dodecahedron and a small cube. In that world, there is something, the small cube, which is not a large dodecahedron, so your translation comes out true. But it shouldn't, since that's *not* a world in which it's not the case that something is a large dodecahedron!

This would work:
~Ex (Large(x) & Dodec(x))
or, equivalently, 
Ax (Large(x) -> ~Dodec(x))

> Ex (Dodec(x) & ~Large(x))

English: "It's not the case that something is a large dodec." Consider a world with two objects, a large dodec and a small dodec. In that world the English sentence is false, but your sentence is true.

Keep in mind that "it's not the case that something is a large dodec" means that there are no large dodecs. (Actually the easiest way to translate it is probably to translate "something is a large dodec" and then add a negation sign in front!)


9.17.2

> Ax (Cube(x) & Small(x) & RightOf(x, a))

English: "Each small cube is to the right of a." Universal affirmative, so form is Ax (S(x) -> P(x)). In this case S(x) is "x is a small cube" and P(x) is "x is to the right of a" so we get: "Ax ((Cube(x) & Small(x)) -> RightOf(x,a))".


9.17.15

> Ax ((SameCol(x, a) v SameCol(x, b)) -> ~Cube(x))

There's one odd thing and one actual problem with this translation.

The odd thing is that whereas the English says something *is* a cube if and only if it's *not* in the same column as a or b, your translation has something *not* being a cube on one side, and *being* in the same column as a or b on the other. Actually switching the negation doesn't hurt anything, since a conditional is equivalent to its contrapositive, but it's a good policy to stay as close as you can to the English.

The problem is that the English is "if and only if"; you just have the "only if" part. You need a biconditional rather than a conditional.


Last update: October 5, 2009. 
Curtis Brown  |  Symbolic Logic   |  Philosophy Department  |   Trinity University
cbrown@trinity.edu