Invalid Logic ArgumentDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/