Philosophy 2340
Symbolic Logic

Advice on Chapter 13 Problems


13.4

> Dr. Brown, I did not include 13.4-if ~ax(Cube x) with the conclusion
> ~ax(Cube x and Small x). I know that it must be true because if all x
> are not Cubes then all x's can not be small cubes.

Right. ~Ax(Cube(x)) means that not everything is a cube (some things could be, but not everything). Not certain whether this is what you meant by "all x are not cubes" since that could also be read as meaning that there are no cubes.

Anyway, I think you're on the right track. Here's one way to think of it: if everything *were* a small cube, then everything would be a cube, contrary to our premise. So it must *not* be the case that everything is a small cube.

Pursuing this informal reasoning, you might try assuming the opposite of the conclusion (i.e. assume Ax(Small(x) & Cube(x))) and then derive a contradiction so you can use negation introduction to derive the final conclusion.

> 13.4, I didn't really understand how to utilize the negation 
> in front of the
> universal quantifier. 

Right -- you can't use the quantifier rules on a formula that's actually a negation. (This is just like all the other rules: you can't use them on subformulas, only on formulas that match the form of the rule. ~Ax Cube(x) is not a universal but a negation, so you can't use universal elimination on it.

(So what to do? Try negation introduction.)


13.6

> In 13.6, I am very close, but as you 
> can probably see
> I need a Tet(a) which I had a hard time proving within one of 
> the subproofs.

That's because you can't prove it inside that subproof! At least not directly. The idea of doing disjunction elimination is not a bad one. In one of the subproofs you can prove Tet(a) and use it to get BackOf(a,c) which is what you really want. In the other subproof, you can derive a contradiction and then use contradiction elimination to get what you want.


13.8

> 13.8 I'm not sure how I can use the rules to turn a 
> conditional statement into a conjunction, so that's a big 
> sticking point. 

This may not be quite the right way to think about it -- you can't really turn a conditional into a conjunction (P->Q doesn't imply P & Q). But the conjunction you want is actually the antecedent of a conditional. (If this were propositional logic, you'd need to show that from P->(Q->R) it follows that (P & Q)->R.) It's a little trickier with the quantifiers thrown in, but basically it'll be like proving any other conditional -- you just have to show that *if* P & Q, then R.


13.9

> On 13.9 I keep getting close to the answer, 
> but I can never get the right variables in the solution. I 
> know I'm close on both of these, but I'll just have to talk 
> to you to figure out why I'm not getting them.

Don't forget the two hints! (1) keep in mind that the arguments may not be valid, in which case you need a counterexample rather than a proof; (2) a universally quantified conditional Ax(P(x)->Q(x)) is true if nothing is P (since then the antecedent isn't true for any objects, and a conditional with a false antecedent is true).

13.30

"I am having some trouble starting 13.30.  I remembered you saying that after you apply universal elimination, you are left with a conditional.  However, I am a bit confused as to if I am to begin a subproof like straight propositional logic or if I should select a constant and get rid of the quantifier."
 
Neither one, actually! You can't use existential elimination on the sentence you got in line 3 because it isn't an existential, it's a conditional (although it has an existential antecedent). But there's also no reason to start a subproof. (What rule would it be for? ->Intro uses a subproof, but you aren't trying to prove a conditional. Rather, you already have one.)
 
Here are two hints. (1) In class, I asked whether people wanted to use "max" or "carl" in the universal elimination step, and someone yelled "max," so that's what I used. However, you might want to consider whether "carl" would be a better choice! (2) If you could find a way to prove the antecedent of the conditional on a separate line, then you could use ->Elim.

 

13.31

"For 13.31 I feel like I am in the right direction, but I am stuck and was hoping you could give me a few hints."
 
I'm a little confused by what you're doing in this one. Your line 3 doesn't seem to correspond to any rule we've got!
 
Remember the general strategy we discussed:
 
1. set up EElim and AIntro, if needed
2. do AElim
3. propositional stuff
4. EIntro
5. finish the EElim and AIntro subproofs, if any
 
In this proof, you are trying to prove a universal, so you *will* need AIntro. Also, you have an existential premise, so you will also need EElim. So you need to set up the subproofs for those two rules first. After that you can do AElim.
 
The universal intro subproof is to prove Ax Ey Likes(x, y). So you need to start a subproof with a boxed constant (a, let's say). Nothing else should appear on the line with the constant. The last line of the subproof will be the result of dropping the universal quantifier and replacing all its variables with 'a'. That means the last line of that subproof will be Ey Likes(a, y).
 
The EElim subproof is based on line 2, which says Ex Ay Likes(x, y). So you need to start a subproof in which the first line is a boxed constant (b, say) followed by the result of taking that sentence, dropping the Ex, and replacing the x's with b's. That gives you Ay Likes(b, y). The last line of that subproof will be whatever you're trying to prove at the next level out, which is Ey Likes(a, y).
 
By the time you've got all that set up, you're getting pretty close. (The whole thing took me 10 steps, including the two premises.) After setting up the AIntro and EElim subproofs, it's time to do AELim (twice, being careful about which constants you use for which variables), then one propositional step, then one EIntro step, and then finish off the subproofs you set up at the beginning.


13.40

Note that the premise is not an existential, it's a conjunction. So you can't use existential elimination on it -- you need to do conjunction elimination to get the existential separately, then do ExElim on that.

13.45

This is *very* close, but you're trying to close off two subproofs at the same time -- can't do that, because each subproof works differently. You have an ExElim subproof inside a negation introduction subproof -- need to conclude the existential elimination subproof first (still inside the negation introduction one), then conclude the negation introduction subproof.

Last update: March 27, 2007. 
Curtis Brown  |  Symbolic Logic   |  Philosophy Department  |   Trinity University
cbrown@trinity.edu