Philosophy 0540: Problem Sets

The PDF version of the syllabus also contains the problem sets.

Some Reminders

It is a requirement of the class that all of the problem sets must be completed and submitted for marking. (We'll let you off once, if you do miss one, but you will get no credit for that problem set.) Failure to submit all (but one) of the problem sets will automatically lead to a grade of NC. Please note that the requirement is that the problem sets should be "completed", and by that I mean that one has given them a proper effort. Simply turning in a piece of paper with a few random jottings does not count as completing a problem set.

As with any mathematical subject-matter, it is impossible to learn this material without doing a lot of exercises. The book contains many more exercises than are assigned, and students are encouraged to do additional exercises to improve their understanding of the material. Students are also encouraged to work on the problems together—though, of course, submitted material should be a student's own work.


Problem sets are graded on the following scale:

Do not compare problem set scores across graders. Different graders always have slightly different standards, and there is not really anything to do about that other than to take it into account. Which we will.

The Problem Sets

Problem Set 1

Section IA (pp. 253-5): 1b,d; 2, 3; 4b,d,f,i,k
You should submit problem 1(d), but it will not affect your grade. What we'd like you to do is to think this through and try to find a sentence that is true if 1(d) is false, and false if 1(d) is true. Whatever your answer, please explain why you think it is correct.

Problem Set 2

Section IB (pp. 255-60): 1a; 2a,c; 3a,c,e; 4b,d,f; 6; 7a,c
Problems 13 and 14 are not assigned, but some of you may wish to do them. They are the sorts of 'logic puzzles' that are found on the LSAT. They can be done by schematizing the various assertions and doing a truth-table, but it is probably easier just to think through them.

Section IC (pp. 260-4): 1; 2; 5; 6; Extra Problems 1-2
NOTE: The Extra Problems are NOT optional. They are just problems that are not in the book.
Problem 6 is quite difficult to do completely correctly. If you're struggling with it, then consider some examples, and see if you can find the pattern that lies behind this fact. It's enough just to try to explain the pattern.
If you'd like an extra challenge, then try 7, 8, and 9.

Problem Set 3

Section IIA (pp. 265-7): 1b,d; 2b,d; 3b,d; 4a,d,g
You should definitely do more of these for practice, but you only need to turn those ones in (though you can also turn in more if you want us to look at them).

Section IIB (pp. 267-71): 1a,c; 3b,c,e; 5a,c; 6
For 6, you may use the "method of §25", if you wish, though we will not discuss it. You probably should just give an informal argument (i.e., one in words). Part of what makes this problem tricky is that it is not immediately clear how to show that something implies a disjunction if it doesn't imply one of the disjuncts (which, in this case, it does not). One answer is that, if you want to show that P or Q, then what you can do instead is show that, if ~P, then Q. That is because ~P → Q is just equivalent to P ∨ Q. So, for 6: To show that

∀x(Fx ∧ Gx → Hx) and ∃x(Fx)


~∀x(Gx) ∨ ∃x(Hx)

you can show that they imply

~~∀x(Gx) → ∃x(Hx)


∀x(Gx) → ∃x(Hx)

And to do that, you can show that

∀x(Fx ∧ Gx → Hx) and ∃x(Fx) and ∀x(Gx)

together imply


I.e., we are showing that the original premises together with the antecedent of the conditional we're triyng to prove imply its consequent. That this works is what the general laws I called "the deduction theorem" says: Some schemata imply a conditional if, and only if, those same schemata, together with the antecedent of the conditional, imply its consequent.

So, practically speaking, instead of 6, you can think of yourself simply as doing this problem:

Show that ∀x(Fx ∧ Gx → Hx) and ∃x(Fx) and ∀x(Gx) together imply ∃x(Hx).

Additional practice problems, with solutions, can be found here.

Problem Set 4

Section IIIA (pp. 273-6): 1a,d,g; 2a,c; 3a,c; 5a,b; Extra Problem 3

Additional practice problems, with solutions, can be found here.

Problem Set 5

Section IIIB (pp. 276-81): 1b,d; 2b,d,f

Additional practice problems, with solutions, can be found here.

Problem Set 6

Section IIIB (pp. 276-81): 4a,b; 5; 6; 9; 14

Note: In 4a, the question is asking you to specify predicates of English that have the properties mentioned. For example, suppose we were asked for a predicate that is irreflexive, symmetric, and intransitive. Then "(1)+(2) is odd" would do. This is irreflexive [∀x¬(x+x is odd)], symmetric [∀x∀y(x+y is odd → y+x is odd)], and intransitive [∀x∀y∀z(x+y is odd ∧ y+z is odd → ¬(x+z is odd))]. The first two of these are obvious. For the last, if the antecedent holds, then either y is even, x and z are both odd, or y is odd, or x and z are both even. Either way, x+z is even.

Additional practice problems, with solutions, can be found here.

Problem Set 7

Section IV (pp. 284-8): 1b; 2b; 3b; 4b,d 5a; 7

Optional: Section IIIC (pp. 281-3): 1; 3

Note: For problem (1), the idea is to show, e.g., that you can deduce ¬∀x(Fx) from ∃x(¬Fx) without using CQ, but only using the other rules of the system. Obviously, to show that CQ is dispensible in this sense, you would need to do all the cases. But, for this problem, you need only one; you can choose which to do.

Richard Heck Department of Philosophy Brown University