## PHL310 Valid Reasoning II, past assignments

Past Assignments28 JanuaryHere is a rough 14-week plan for us. Refer back to this if you ever are thinking: what are we doing right now? I changed it very slightly for us, since I figured we should do some more formal set theory later.

Week

1. Natural deduction system: propositional logic (review).

2. Natural deduction system: quantified logic (review).

3. Getting accustomed to multiple quantifiers. Introduction to some basic set theory.

4. Axiomatic system for propositional logic. Proofs with the axiomatic system. Different axiom systems.

5. Mathematical induction. Strengthening. Proving completeness of propositional logic.

6. Axiomatic system for quantified logic.

7. Completeness of quantified logic.

8. Completeness of quantified logic, continued.

9. The Dedekind axioms (AKA the Peano axioms).

10. Axiomatic set theory.

11. Some brief observations about the axiom of choice. Maybe also a very brief collision with recursion theory.

12. Modal logic. Basic systems.

13. Modal logic semantics.

14. Applying modal logic to philosophical problems. Perhaps also: deontic logic, logic of time.

30 JanuaryI'd like to see what system you learned, and where you are at. Five problems, just for me to see. The first two are just translations into some kind of formal logic. For problem 1, use a propositional logic (the smallest things in your language will be sentences). For problem two, use a quantified logic with predicates. For 3-5, generate a proof (a syntactic proof, not a truth table). Use whatever rules or proof methods you remember.

- If both Nemo and Jaws are fish, then neither Patrick nor Mr. Krab are.
- All men are mortal.
- Premises: (P --> Q), (R--> Q), (P v R). Conclusion: Q.
- Conclusion: ((P-->Q) --> (~Q --> ~P)
- Premises: ∀x(Fx --> Gx), ∃xFx. Conclusion: ∃xGx

6 FebruaryHere are some problems to try. Each is in propositional logic. Translate 1-4. Try to do 5 and 6 with a direct proof; 7 and 8 with a conditional proof; and 9 and 10 with an indirect proof.

- Tom will go to London or Paris but not both.
- Tom will go to London if and only if he goes to New York or Newark.
- Tom will go to London only if he doesn't get arrested.
- Tom will go to Paris provided that he doesn't get arrested.
- Premises: (P --> Q), (S ^ ~T), (T v P), (~R --> ~S).

Conclusion: (Q ^ R)- Premises: (~P v Q), (Q --> S), ~S.

Conclusion: ~P- Premises: (~R v T), (T --> Q), ((Q ^ S) <--> V), (V <--> P).

Conclusion: ((R ^ S) --> P).- Conclusion: ((P v Q) --> (~P --> Q))
- Premises: (P --> R), (Q --> S), (~R ^ ~S).

Conclusion: ~(P v Q).- Conclusion: ~(P ^ ~P).
- Extra credit: Conclusion: (~(P ^ Q) <--> (~P v ~Q)).
- Unfair late added question. Can you make a sentence form equivalent to (P --> Q) using P, Q and some combination of ^, v, <-->, and ~?
18 FebruaryReading, and a homework.

Reading: read sections 1.1, 1.2, and 1.3 of Mendelson.

Homework, due at the beginning of class: some natural deduction system quantifier review. Translate 1-8. 1-4 are standard quantifier translation forms; 5-8 require multiple quantifiers. Give proofs for problems 9-12. 9 and 10 can be done directly. 11 and 12 require a universal derivation.

- All cats are mammals.
- Some mammals lay eggs.
- Some mammals do not lay eggs.
- No mammals reproduce asexually.
- Some number is bigger than some number.
- Every number is bigger than every number.
- Every number is bigger than some number.
- Some number is bigger than every number.
- Premise: ¬Fa, ∀x(Fx v Gx), ∀y(Gy --> Hy). Conclusion: Ha.
- Premise: ∃x(Jx ^ Kx), ∀x(Jx --> Lx), ∀x(Kx --> Mx). Conclusion: ∃x(Lx ^ Mx).
- Premises: ∀x(Fx --> Gx), ∀x(Gx --> Hx) Conclusion: ∀x(Fx --> Hx)
- Premises: ∀x(Fx --> Gx), ∀x(Hx --> Ix) Conclusion: ∀x((Fx ^ Hx) --> (Gx ^ Ix))

20 FebruaryRead section 1.4 of Mendelson. You might enjoy 1.5 and 1.6, but they are a bit advanced.

22, 25 FebruaryWe will (1) look at some proofs in the axiomatic system; and (2) introduce induction! Induction is likely something new for most of us, so we'll go slowly, and try it out in a few cases.

Here are two things that I promised to post. First, my proof using our axiom system, without any lemmas, of (¬¬B --> B).And here is the first induction example we did in class.

- (¬B --> ¬¬B) --> ((¬B-->¬B) --> B) A3
- (¬B --> ((¬B --> ¬B) --> ¬B)) --> ((¬B --> (¬B --> ¬B)) --> (¬B --> ¬B)) A2
- (¬B --> ((¬B --> ¬B) --> ¬B)) A1
- ((¬B --> (¬B --> ¬B)) --> (¬B --> ¬B)) MP 2, 3
- (¬B --> (¬B --> ¬B) A1
- (¬B --> ¬B) MP 4, 5
- ((¬B-->¬¬B)-->((¬B-->¬B) --> B)) --> (((¬B-->¬¬B)-->(¬B-->¬B)) --> ((¬B -->¬¬B)-->B)) A2
- (((¬B --> ¬¬B)--> (¬B-->¬B)) --> ((¬B --> ¬¬B)-->B)) MP1, 7
- ((¬B-->¬B) --> ((¬B --> ¬¬B) --> (¬B-->¬B))) A1
- (¬B --> ¬¬B) --> (¬B-->¬B)) MP 6, 9
- ((¬B --> ¬¬B)-->B)) MP 8, 10
- (¬¬B --> (¬B --> ¬¬B)) A1
- (¬¬B --> ((¬B --> ¬¬B)--> B)) --> ((¬¬B -->(¬B --> ¬¬B)) --> (¬¬B -->B)) A2
- ((¬B --> ¬¬B)-->B) --> (¬¬B --> ((¬B --> ¬¬B)-->B)) A1
- (¬¬B --> ((¬B --> ¬¬B)-->B)))) MP 11, 14
- ((¬¬B -->(¬B --> ¬¬B)) --> (¬¬B -->B)) MP 15, 13
- (¬¬B --> B) MP 12, 16
Show that the sum of the first n odd numbers is n^{2}.

That is, show: 1 + 3 + 5 + ... (2n - 1) = n^{2}.

BASE CASE:

The first odd number, which is 1. The sum of all odd numbers up to 1 is 1, and 1 = 1^{2}.

INDUCTION HYPOTHESIS:

The sum up to the n^{th}odd number, which has value (2n - 1), is n^{2}.

SHOW:

The (n + 1)th case is the sum up to the odd number after the nth odd number.

Since the n^{th}odd number has value (2n - 1), the odd number after that has value (2n - 1 + 2). So, we need to show that (n+1)^{2}= n^{2}+ (2n - 1 + 2).

Note then that, (n+1)^{2}= (n+1)(n+1) = n^{2}+ 2n + 1.

Note that n^{2}+ (2n - 1 + 2) = n^{2}+ 2n + 1, so (n+1)^{2}= n^{2}+ 2n + 1.

But then, since n^{2}+ 2n + 1 = n^{2}+ 2n + 1, it follows that (n+1)^{2}= n^{2}+ (2n - 1 + 2).

CONCLUSION:

The sum of the first n odd numbers is n^{2}.

26 FebruaryIn case anyone wants to review problems in the natural deduction system: Extra review session, 12:30 p.m. - 1:30 p.m., in my office. Here are some sample problems for the natural deduction system. Do as many as you like for practice. I'll review any or even all of the problems below if you want. If you can't make it you could come to me another time with any of these problems.

- Premises: (¬¬P --> S), P. Show: ¬¬S.
- Premises: ¬(T --> R), ((P --> V) --> (T --> R)). Show: ¬(P --> V).
- Premises: (¬P --> V), ¬V, (P --> Q). Show: Q.
- Premises: (¬(R --> S) --> V), ((R --> S) --> Q), ¬Q. Show: V.
- Premises: (P --> R), (R --> Q). Show: (P --> Q).
- Premises: (S --> R), (P --> S). Show: (¬R --> ¬P).
- Premises: (¬P --> ¬S), S. Show: P.
- Premises: ¬(P --> Q), ((R --> T) --> (P --> Q)). Show: ¬(R --> T).
- Premises: (P --> Q), (Q --> R), (R --> S), (S --> T), ¬¬P. Show: T.
- Premises: (¬(R --> S) --> V), ((R --> S) --> Q), ¬Q. Show: V.
- Premises: (P --> R), (T --> P), (S --> T), ¬R. Show: ¬S.
- Premises: (R --> Q), (Q --> S), (P --> R). Show: (P --> S).
- Premises: (P --> S), (Q --> P), (¬R --> Q). Show: (¬S --> R).
- Premises: ((T v R) <--> S), (S ^ ¬Q), (T <--> Q). Show: R.
- Premises: (P <--> R), (R ^ ¬S), (S v T). Show: ((Q v P) ^ T).
- Show: (¬Q --> ¬(P ^ Q))
- Show: (Q --> R) --> ((P --> Q) --> (P --> R))
- Show: ¬(P --> Q) <--> (P ^ ¬Q)
- Show: (P ^ Q) <--> ¬(¬P v ¬Q)
- Premises: (P ^ R), (P --> S), (Q --> ¬(S ^ R)). Show: ¬Q
- Show: ((P ^ ¬Q) --> ¬(P --> Q))
- Show: (¬(R --> T) --> ¬T)
- Premises: (P <--> R), (R <--> S), (S <--> Q). Show: (P <--> Q).
- Make the truth table for the following sentence: (¬(P v Q) <--> (¬P ^ ¬Q)).
- Make the truth table for the following sentence: (¬(P ^ Q) <--> (¬P v ¬Q)).
- Make a truth table for the following sentence: ((¬P --> Q) <--> ¬R).
- Make a truth table for the following sentence: ((R ^ S) <--> (P v S)).
- Show the following argument is not valid using a truth table. Premises: (P v Q), P. Conclusion: ¬Q.
- Show the following argument is not valid using a truth table. Premises: (¬R --> S). Conclusion: ¬R.
- Show the following argument is not valid using a truth table. Premises: ¬(T --> V). Conclusion: ¬T.
- Show the following argument is not valid using a truth table. Premises: ¬(R^S). Conclusion: ¬R.
- Show the following argument is not valid using a truth table. Show that this is not a tautlogy: ((P ^ Q) <--> (P v Q))
- Premises: ∀x(Fx --> Gx), FC. Conclusion: GC.
- Premises: ∀x(Fx <--> Gx), ∀x(Gx <--> Hx), Fa. Conclusion: Ha.
- Premises: ∀x(Gx v Hx), ¬Hb. Conclusion: ∃xGx.
- Premises: ∀x(Fx --> (Gx ^ P)), ∀xFx. Conclusion: P.
- Premises: ∀x(Q v (Gx v Hx)), ∀x¬Gx, ¬Q. Conclusion: HC.
- Show: ∀z(Fz <--> Gz). Premises: ∀x(Fx <--> Hx), ∀y(Hy <--> Gy).
- Show: VxFx. Premises: Vx(Gx ^ Hx), ∀x(Gx --> Jx), ∀x((Hx ^ Jx) --> Fx).
- Show: Vx(Gx ^ ¬Hx). Premises: ∃x¬Hx, ∀x(Gx v Hx).
- Show: ∀x(Fx --> Gx) --> (∀xFx --> ∀xGx)
- Show: ¬∀xFx <--> ∃x¬Fx

1 MarchIn class:We'll review the proof of the deduction theorem, moving slowly through it to ensure we see how induction got used.

Homework due:Proofs in the axiomatic system. First two practice problems for your own edification; I won't ask you to hand them in. (1) Make truth tables for our three axioms to convince yourself that they are tautologies. (2) Using just substitution to find instances of our three axioms, and modus ponens, show:These problems are actually proven in your book. Do the following: try them on your own; then look at the book's proof; then see if you can fill in the blanks (that is, show the steps that lean on lemmas). Next,

- (¬¬B --> B)
- (¬B --> ¬A) --> (A --> B)
fourproblems I will ask you to hand in. First, show the following:Remember that "|-

- (A --> (B --> C)), (A --> B) |-
_{L}(A --> C)- |-
_{L}((A --> B) --> ((C --> A) --> (C --> B)))- |-
_{L}(B --> (¬B --> C))_{L}" just means provable in our system L. The stuff on the left is premises; the stuff on the right is your conclusion.

I'll allow you the chain rule to make things simpler. This is known as corollary 1.10a in your book. For problem 3, I'll allow the deduction theorem (this means that you can do the equivalent of a conditional derivation). Also, for problem three, note how like problem 1.11c this is. If you're miserably stumped on problem 2, then use the deduction theorem.

Here is problem4: It's the "hello world" of induction problems. Solutions are ubiquitous, so I'm asking you to use the honor system and give it a try all on your own. Suppose we have some basic arithmetic available. Then an equation that I believe Gauss proved when he was a kid will be a theorem: 1 + 2 + 3 + ... + n = (n^{2}+n) /2. Show this using mathematical induction. It can be done, by the way, with weak induction.

Hints: remember that 1 is your base case; and also, factor (n^{2}+n) /2 to (n*(n+1))/2; let n be your induction case, and then you'll be proving this is true for n+1. More hints: What will your proof step be like? Well, note that you get to assume that 1+2+3... +n = (n*(n+1))/2. Now, for sake of illustration, let's suppose the next number is z. Then we can see that the sum 1+2+3+... +n+z is (n*(n+1))/2 + z, assuming our induction hypothesis. Thus, you've proved the n+1th case if you can show that (z*(z+1))/2 = (n*(n+1))/2 + z. OK, you can do that if you can find a way to say z that will allow you to solve. And that's obvious: what is z, after all, in relation to n?

In class I promised my answers to problem 2 and answers to problem 4.

March 8Let's do some practice problems. We will go over these on Wednesday, and have our quiz Friday; that way we have a lot of time to prepare. Let me know if that schedule works. I'll collect and grade these, as incentive.

For problem 2, see how we translate "^", then you may use theorem 1.11c (theorem 1.11c is: (¬B --> (B --> C)); remember that you can substitute a negation sentence for any of those variables B or C). Problem 3 is an induction problem; like our last one, it is a very common one. Please do it without looking up the answer.Some hints:

- Show: (B --> C), ¬C |- ¬B
- Show: (B ^ C) |- B
- Show: 1
^{2}+ 2^{2}+ 3^{2}+ ... n^{2}= n(n+1)(2n+1)/6My solutions are here. They are not the only way to prove these results.

- For problem 1, I found the easiest way to prove it is to use this instance of axiom 3: (¬¬B --> ¬C) --> ((¬¬B --> C) --> ¬B)), and also theorem 1.11a (which says (¬¬B --> B)), and also to use deduction theorem in the middle of my proof to prove a conditional I needed along the way; I also used axiom 1 once.
- For problem 2, I used the following instance of axiom 3: (¬B --> ¬(B --> ¬C)) --> ((¬B --> (B --> ¬C) --> B), and I used theorem 1.11c (which says (¬B --> (B --> C)); remember, when using a theorem, that it's like an axiom, and you can replace B or C with any wff as long as you replace it everywhere in the theorem consistently; so, for example, you could replace C with ¬C.) One instance of A1, a few modus ponens, et voila.
- Problem 3 requires a bit of math. But, if you show me how you set up the problem, I will give you most of the credit, even if you fail to get it all. But here's a big hint: For the induction step, the easiest way to do it is to factor out (n+1), and then you can show both your own reasoning and the equation yield the answer (n+1)(2n
^{2}+ 7n + 6)/6.

March 15Test. Simple proofs with axioms. Using the weird lemma to prove completeness. Meaning of: completeness, consistency, soundness. Applying and using these concepts.

The mean on the test was 33.2. I posted very approximate midterm grades; they are rough because they count the homeworks too much, and a few of you missed one or two homeworks and that dragged down the grade. But we'll do many more homeworks, and more tests, so this will average out more accurately in the weeks to come. We'll review the test when we return.

March 18Over break, read 2.1-2.5.

March 27We'll review very briefly what we've learned with the propositional logic. Then, we'll turn to a quantified logic. There's much that we must learn! I think we'll handle the proofs more lightly, so that we have time in the rest of April to get to fun stuff like modal logic, the Peano Axioms, maybe some set theory or recursion theory.

I don't want to post homework until we've introduced the new axioms. However, give a try to sections 2.1-2.5. You'll find it hard going, but we'll familiarize ourselves with that stuff, and that will help us learn the material together. I'll post a homework next week.

April 3Homework. Complete the following. They are informal, because we haven't specified a language, but you can manage them I think.

- Consider a model concerned with numbers. For each of the following, give an example of a relations that satisfies the predicate. The relation will be an ordered n-tuple for an arity n predicate, written between "()", which we take to mean ordered set. So, for example, (3, 6) would satisfy the predicate "___ is half the size of ___."
a. >

b <

c. =

d. ... is the next largest prime number after ....- Consider a model concerned with animals, including humans. For the following, give an example of a relation that satisfies the predicate. The relation will be an ordered n-tuple, written between "()", which we take to mean ordered set. You may need to fabricate, for the sake of the exercise, proper names for particular animals; just tell us what kind of animal the individual referred to is.
a. ... is a mammal.

b. ... is of a species that shares a common ancestor with ....

c. ... is of a species that is extinct.

d. ... and ... are the parents of ....

- Consider a model concerned with numbers. For each of the following, give an example of the n-tuple relation that would go into the function, and then the example of the one thing that would come out of the function. For example, (6, 3) in the function "+" would yield (9).
a. +

b. x

c. -

d. / [divided by]- Consider a model concerned with animals, including humans. For each of the following, give an example of the n-tuple relation that would go into the function, and then the example of the one thing that would come out of the function. For example, (G. W. Bush) put into the function "father of" would yield (G. H. W. Bush).
a. "biological mother of ..."

b. "biological father of ..."

c. "biological paternal grandfather of ..."

d. "biological maternal grandmother of ..."

10 AprilI think Andy was right to suggest we dig in a little onto the object language, so that we get a good sense of what the axioms are empowering us to do. Using just the axioms and gen and the (new) deduction theorem, prove the following theorems. I wouldn't spend time trying to do them without the deduction theorem; it would be hard and time consuming. I think you you may find you don't need axiom 5 for these.

- ∀x
_{1}(B(x_{1}) --> C(x_{1})) --> (∀x_{1}B(x_{1}) --> ∀x_{1}C(x_{1}))- ∀x
_{1}(¬C(x_{1}) --> ¬B(x_{1})) --> ∀x_{1}(B(x_{1}) --> C(x_{1}))

10, 12 AprilWe're going to prove completeness of FOL. Here are the parts we'll prove, and what we'll just assume. Take a look at these and study a bit the path we are going to follow.I would read and review chapter 2 up to the completeness proof (and the completeness and soundness proof named "Godel's Completeness Proof").

- Prove deduction theorem
- Assume consistency observation
- Prove proposition 2.1
- Prove lemma 2.12
- Assume Lindenbaum's lemma
- Assume lemma 2.15
- Prove proposition 2.17
- Prove completeness

After this, we'll peek at some set theory, then Peano arithmetic, and then do some modal logic. The modal logic is really fun because we can engage in some speculative metaphysics about time, possibility, necessity. I'll even tell you about zombies.

15 AprilLet's start our brief tour of axiomatic set theory. Read chapter 4, section 4.1. If you feel ambitious check out sections 4.2 and 4.3. We'll pick sparsely from those.

Here are our goals for our review of the set theory:

- explore how to create the set theory out of axioms;
- learn some of the basic concepts, like power set, equinumerosity, and ordering;
- peek at how you can make the numbers out of sets;
- discuss two of the controversies (constructivism to avoid the Set Theory Antinomy; and also the issue of the axiom of choice).

17 AprilQuest.

19 AprilIn class: review of proofs in K; review of our context (we are extending K and L for the next three topics); some examples of semi-formal set theory proofs; Cantor's Theorem.

22 AprilCantor's theorem. Constructing the natural numbers.

24 AprilHomework really due!

Here are some helpful things. here are my inelegant proofs of the last two problems in a pdf.

For this homework, you need to make clever use of A3 often. Here are some hints.

For 1, note the following is a version of A3: (¬¬Fx --> ¬Gx) --> ((¬¬Fx --> Gx) --> ¬Fx). You should take note of theorem 1.11a and corollary 1.10a.

For 2, note the following instance of axiom 3: (¬Fx --> ¬(Fx --> ¬Gx)) --> ((¬Fx --> (Fx --> ¬Gx)) --> Fx). Consider also theorem 1.11c.

For 3, consider again 1.11c and also A3, such as instance (¬Gx --> ¬(Fx --> ¬Gx)) --> ((¬Gx --> (Fx --> ¬Gx)) --> Gx).

For 4, first note the following instance of lemma 1.11f: ∀x(Fx-->Gx) --> (¬¬∀x(Fx-->Hx) --> ¬(∀x(Fx-->Gx) --> ¬∀x(Fx-->Hx)). To get the parts you need, consider your two earlier proofs, and consider lemma 1.11b and corollary 1.10a.

You can prove the set theory claims either informally or formally. Use for these the natural deduction theorem tricks of every kind that you know. Remember your definitions of subset, intersection, union, and identity. To prove an identity formally, you can prove both directions of the biconditional for the arbitrary instance.

In class: How do we feel about these infinite sets? And: a super high level discussion of Choice. Looking at the axioms for Peano Arithmetic.

26 AprilRead section 3.1. There's some weird/difficult stuff there, but skip to the parts you can figure out.

Let's play with S to get a feel for it. Using just the axioms of S (page 150 in the latest edition of our book) prove thatWe allow free substition of identicals throughout your proofs, so you can always replace a term with an identical term. What that means is that you can have proofs like the following proof, given as a simplistic example, that 1+0=0+1:

- 1+1=2 (In the object language, this means you'll prove that 0'+0'=0''. Note also that parenthesis that surround a function disappear when we replace the function with its referent, so don't be confused by this. (That is, suppose that n+m is 0, then (n+m)' is 0', not (0)'. Besides, we will always take (n)' to be equivalent to n' once you put in for n the actual term in question. So parentheses are not changing meaning.) (Hint: look at S5 and S6.)
- 2 x 1 = 2 (Hint: look at S7 and S8; the dot is multiply.)
1. 0' + 0 = 0' ....... Axiom S5

2. 0 + 0' = (0 + 0)' ...... Axiom S6

3. 0 + 0 = 0 ...... Axiom S5

4. 0 + 0' = 0' ...... substitution of identicals, 2, 3

5. 0' + 0 = 0 + 0' ...... substitution of identicals, 1, 4

29 AprilMore on number theory! Start of modal logic.

1 MayModal logic systems. Leibniz semantics and its challenges. Read appendix B of our book.

3 MayHomework. Due at the beginning of class. Use any theorem in our book to prove 1 (this takes a few steps, and it is helpful to use some of the sub-propositions in proposition 3.2). Use any natural deduction rules you like for problems 2 and 3, although you are permitted as axioms only the axioms of L and M.

- Prove using S alone: 2 x 3 = 4 + 2
- Prove using M alone: A --> <>A
- Prove using M alone: []A --> <>A

8 MayHomework. Due at the beginning of class. Try these early and then let me know if you need hints. This one will count a lot because there are so many questions. You can use any rules of propositional logic, but not any axioms or corollaries of modal logic other than the axioms specified in the system you are to use and any corollary you proved in the last homework. Remember that all our systems contain M, so all of them include m1 and m2 as axioms.

For these, it is helpful to use the equivalence principle. If you know two things are equivalent, we will now let you replace one with the other eveninsidesentences. So, because you know ~~B <--> B, for any B, then you can add or remove "~~" anywhere you want in a sentence, even inside the sentence.

- Prove in S4 (that is, using only m1, m2, m4, necessitation): []A <--> [][]A
- Prove in S4 (that is, using only m1, m2, m4, necessitation): <>A <--> <><>A
- Prove in S5 (that is, using only m1, m2, m5, necessitation): []A <--> <>[]A
- Prove in S5 (that is, using only m1, m2, m5, necessitation): <>A <--> []<>A
- Show S5 is equivalent to the combination of Brouwer and S4 (this means, show that from the axioms of Brouwer ( m1, m2, m3) and S4 (m4) you can derive the axioms of S5 (m5); and then show that from the axioms of S5 (m1, m2, m5) you can derive the axioms of Brouwer and S4 (m3 and m4)).
- Given Kripke's semantics (and his claims about the required accessibility relation), why might S5 make a bad logic of time (given our usual assumptions and experience of time)? We haven't talked about time yet, but look at your handout and the possible temporal interpretation of [] and <>, and think through the implications. (There is an issue here about whether propositions must include a time stamp, but if that worries you, assume that some propositions in your language do not.)
- If we use our modal operators for a deontic logic (a logic of ethics), and interpret [] as meaning "it should be the case that", and we interpret <> as meaning "it is permissible that", why is M probably not a good logical system to use as our deontic logic? We haven't talked about deontic logic yet, but look at your handout and the possible deontic interpretation of [] and <> and think through the implications.

10 MayReview of our last homework. And back to Plato's heaven: what has logic revealed for us? Where are those infinite sets, anyways?

13 MayOffice hours 9:00 -10:00 a.m. and 1:30 -- 3:00 p.m.