FAQ List

Chapters: 1 2 3 4 5 6 7 8 9

 

General Questions:

Why does nothing happen when a quiz, resultsTrack, or something else is supposed to be displayed?

You need to make sure your browser allows "pop-up" windows. Some people have "pop-up blockers". Make sure you set these to let the Café open new windows.

How come nothing happens when I try to do a quiz or exercise? When I try to run a Flash presentation?

You may need to set your browser to enable "JavaScript". And you need to download and install Flash.

Chapter One:

What is logical possibility? Why is it important?

We used the notions of possibility and impossibility to define the notion of a valid argument. And we'll continue to use these notions in other definitions. So, it's important to get clear on what they mean. In a nutshell, logical possibility is what might have happened in some possible world, about how things could have been (even if actual matters that have become settled now preclude it).

1.4ex III What is the deal with informal proofs that require a contradiction?

Some of the informal proofs require that you make an assumption for contradiction. We'll talk more about these in class. But the idea is that you can prove that something couldn't be true by showing it leads to contradiction. For example, if you assume that 7x (3 + 4) = (7x3) + 4, then you'll notice that this assumption means that 7 x 7 = 21 + 4 and so leads to the crazy conclusion that 49 = 25. This contradicts the obvious claim that 49 is unequal to 25. SO, THE ORIGINAL ASSUMPTION THAT 7x (3 + 4) = (7x3) + 4 MUST BE WRONG!

1.4ex III 2:

> The proofs that use a contradiction confuse me because it seems like the statement you are trying to prove true is false.

It's the other way around. You ASSUME one statement is false, and show that it can't be false. Because it leads to a contradiction to assume it false, it follows that the statement is true.

>One of them says: "If the conclusion of an argument is >logically true, then the argument is valid."

Suppose the argument is INvalid. Then it would be possible for its premises to be true and conclusion false. But that would mean that the conclusion could be false. Contradicting our supposition that the conclusion is logically true.

>I can sort of
>follow the proof and it makes a little sense, but I still
>think the statement is false because you can have an
>argument that is logically true but that is not worded in a way that makes it valid, can't you?

(BTW, "Logically true" is defined only for sentences, not arguments.)

Anyway, this is a tricky one. But given the way we've defined "valid", an argument is automatically valid if it's got a logically true conclusion. YOU MIGHT THINK THAT THIS MEANS THAT THE DEFINITION WE'VE GIVEN, THEN, IS A FUNNY ONE. THAT'S A GOOD THOUGHT...WE'LL TALK ABOUT THAT.

Chapter Three:

What's with 3.1ex III and 3.2ex V?

You will find some of the truth table exercises allow you to enter your own sentences. It's probably good to do some of these. (). But these are optional. In fact, you'll see more of these in chapter 3 and again in chapter 5. There is no resultsTrack for them and I'll not be checking them in any way.

Chapter Four:

4.1ex I #6 How do you symbolize "unless"?

One of the tricks to learn is with "unless". If I say "we win unless they score", then this means "we win if they don't score" and the symbolization should be something like '~S>W'. BUT this is logically equivalent to 'WvS' and so you can always simply symbolize "unless" with the wedge. Does this make any sense? See tutorial 2.4

4.3ex all: How do we know that an argument is invalid? It seems extreme that one row can show an argument is invalid.

I think the lesson to learn is that if there is even one possible way for the conclusion to be false (while prems are true) then the argument is INVALID.

So, if there is one row making the premises true and conclusion false the argument is invalid...end of story! OK?

Chapter Five:

5.5ex III #3 This exercise is more difficult. It seems impossible? When should I make assumptions, I just can't tell?

Yea, it does get harder with 5.5 ex III. So I wouldn't worry about it too much. Better to look ahead to tutorial 5.6 (you'll notice that this exercise tells you that you may have to).  

So, my first thought is to make sure 5.5 ex I and II make sense. Then study for the exam (5.1-5.3). And only if that's going well, should you go on to look at the next tutorial.

Here are some hints specific to these problems. Only make assumptions under two conditions: 1) you have a goal sentence that has main connective horseshoe; in that case assume it's antecedent. 2) You have a goal sentence, ~P with main connective tilde. In that case, assume it's negate, P. (We'll say more about this in 5.6.)

1  
2  
3  
4  
5 (A&B)>(~J>B)???

Here you have a goal sentence (=what you are asked to derive) which is '(A&B)>(~J>B)'.

So, make an assumption:

Assumption 1 ....what if A&B
  2    
  3    
  4    
  5 (A&B)>(~J>B)???  

Now you know what you need at the end of your subderiation: the consequent '~J>B' of your goal; this TOO has main connective horseshoe, so you need a SECOND assumption.
Assumption 1 ....what if A&B  
Assumption 2 ....then... ....what if ~J
1 &E 3 ....then... ....then... ???
2-3 >I 4 ....then... ~J>B???  
1-4 >I 5 (A&B)>(~J>B)???    
Now you can do the rest:
Assumption 1 ....what if A&B  
Assumption 2 ....then... ....what if ~J
1 &E 3 ....then... ....then... B
2-3 >I 4 ....then... ~J>B  
1-4 >I 5 (A&B)>(~J>B)    

 

5.6ex III #5 I'm having a hard time trying to prove ~A&~B. I think what I have to do is ~I but there is no way for me to prove that in a subderivation, considering the premise is ~(~A&~B). Can you give me a hint?

You'll need sub-sub derivations for this one. First assume ~(AvB) at line 2. Then (in order to contradict line 1) try to prove ~A&~B by proving first ~A. So, assume A!

5.6ex IV About Assumptions: I was going through the homework and the mock exam and I'm confused on when to use assumptions and when to not.

Here's my rule of thumb: NEVER make an assumption unless you EITHER 1) have a goal sentence of the form P>Q then you may want to assume P for later use of >I OR 2) you have a goal sentence like ~R where you need to assume the opposite, here R for later use of ~I or ~E.

So, the only time you should ever make an assumption is when you plan to use >I or one of the two negation rules.

Once you have the short-cut rules of 5.7, you'll need to do even fewer assumptions. Do remember that it's better not to make assumptions because these start subderivations. It's best to avoid subderivations when possible.

 

Chapter Six:

6.1ex IV #6: Everyone scored as well on the LSAT's as Agnes. Should this be symbolized as (^x)Sxa or (^x)Sax or does it matter?

It does matter. You need to write this one as '(^x)Sxa' because that says that each person x scores at least as well as Agnes: Sxa. But 'Sax' would say that Agnes scores at least as well as x.

6.3ex IV: In a sentence like '(^x)(^y)Sxa>Lyx)', why don't we remove both the quantifiers and substitute in for all the variables?

The definition of a substitution instance is just asks you to remove the one quantifier which is the main connective (it's always on the far left) and replace every occurrence of that quantifier's variable with any name (you pick one). We'll need to think about quantifiers one at a time in the chapters that follow.

Chapter Eight:

8.2 and 8.3: I'm having promblems with some of the problems in chapter 8...

> 1. I don't know what to do when there is a tilda in front of
> an entire premise.

 
Yes this can be hard. You can use the "short-cut" rules in some cases. But in many you won't have the appropriate short-cut until you get to 8.4 and can use QN.
 
So, in the mean time, you'll often need to reiterate the negated sentence as part of a ~E or ~I subderivation. See below, 8.2, for a first example.

> 2. I don't know when to make assumptions or what they should
> be(other than with horseshoes as main connectives and
> negating the goal).
> Here are a few problems that I'm having a lot of trouble
> with...
> 8.2ex1#5
 
At line 8, you'll need ~Pa (where 'a' is arbitrary so that you can use VI to prove line 9, yes?) OK, so to prove line 8, you'll do...what? Well you can think of ~I to prove ~Pa, so assume Pa. With a little thinking you'll see how to prove the required contradiction. (Hint: 1 R will be part of it.)

> 8.2ex2#5
 
This one is similar. The goal sentence is universally quantified. So, the line above it should be the substitution instance.

> 8.2ex4#3
 
First assume the antencdent of your goal sentence: (3y)By. Then make a second assumption to set up for 3E: Assume Bt. Then do the usual subderivation for 3E. (Hint: the last line of the sub-sub derivation must be the same as the last line of the subdervation = (3x)Bx.

> 8.3ex2#2&4
 
(If you can't read the following, I'll send this in another format.)
 
Here's 2:
 
 
Premise   1   ~(^x)Px    
Assumption   2   ....what if ~(%x)~Px  
Assumption   3   ....then... ....what if ~Pa
3 %I   4   ....then... ....then... (%x)~Px
2 R   5   ....then... ....then... ~(%x)~Px
3-5 ~E   6   ....then... Pa  
6 ^I   7   ....then... (^x)Px  
1 R   8   ....then... ~(^x)Px  
2-8 ~E   9   (%x)~Px    

Of course, once you reach 8.4 you'll have a "short-cut" rule which allows you to move in one step from ~(^x)Px to (%x)~Px. That's QN.

And here's 4:

Premise   1   ~Ja  
Premise   2   ~(^x)(Gxa>(%y)~Jy)v(^x)Gbx  
Assumption   3   ....what if Gma
1 %I   4   ....then... (%y)~Jy
3-4 >I   5   Gma>(%y)~Jy  
5 ^I   6   (^x)(Gxa>(%y)~Jy)  
6 DN   7   ~~(^x)(Gxa>(%y)~Jy)  
2,7 DS   8   (^x)Gbx  
8 ^E   9   Gba