Invalid Logic Argument

Date: 9/9/96 at 19:44:13
From: Anonymous
Subject: Invalid Logic Argument 

Dr. Math,

Show that the following argument is not valid:

[p=>r],[(not r) or s], [q or (not s)], [p and (not t)] yields [q=>t]

My logic was that if I want the result to be true and then have a 
false premise, then the argument would be invalid.  Because I was 
trying to make everthing true, I started with the conjunction to 
simplify things a bit.

I started with the statement [p and (not t)].  Both [p] and [not t] 
must be true for this premise to be true.

When [not t] is true [t] is false.

The only way for the conclusion [q=>t] to be true while [t] is false 
is for [q] to be false.

If [q]is false and I want [q or (not s)] to be true, then {not s} 
must be true, making [s] false.

If [s] is false and I want [(not r) or s] to be true, then [not r] 
must be true, making [r] false.

If [r] is false and I want [p=>r] to be true, then [p] must also be 
false, BUT in the first step [p] is true.  [p] cannot be both true and 
false; therefore the argument is not valid.


  My instructor marked off the whole problem because I did not make 
[q=>t] false as my first step. I would like to know if I am correct.  

Thanks a whole bunch in advance,  Kara Archer

Date: 9/10/96 at 1:38:30
From: Doctor Mike
Subject: Re: Invalid Logic Argument 

Hello Kara,

You obviously understand quite a bit about this. The logical 
argument in use here is "Reducto ad adsurdum" (please excuse my Latin 
spelling if that is not exactly right), or in English Reduce to 
Absurdity.  That is, assume the opposite of what you want to prove, 
then logically derive an impossibility, and conclude that assumption 
is false.  

If the instructor does not actually see a step of the form "Assume 
...... is ....." then you have not shown beyond a shadow of a doubt 
that you fully understand it.  This is appropriate at the stage of an 
instructor testing comprehension of a new and tricky concept.  You had 
all the details, but they were not tied up in a nice neat package.  
Neat packages are easier to grade. 
  
My chosen way to show I understand this logic concept would be:
 
1. I want to show this is NOT a valid argument, so I will assume
   that it IS a valid argument.
2. Therefore if all 4 premises are true, then q=>t is also true.
3. Exactly your logic, resulting in (p and (not p)), a contradiction.
4. My explicit assumption in (1) can't be true, so the argument
   is invalid.
  
The instructor can see all he/she needs to in order to give full
credit and everybody is happy.  

By the way, there is a direct proof. If you just assume the 4 
premises, then you can conclude that t is false and the others are all 
true.  Then the conclusion q=>t could not follow since it would be 
true=>false.  
  
I hope this helps out. 

-Doctor Mike,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/