Logic Midterm Confusion?

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Nimmy
 
Reply Tue 14 Apr, 2009 08:14 pm
So, we got our second logic midterms of the semester back for our Introduction to Logic class at my school back, and I was disappointed with my score. I did really well on the first midterm, and I thought I knew what was going on so seeing a C+ really threw me for a loop. Can you guys help me understand why what I answered was incorrect?
Our midterms are short, 3 symbolizations and 3 proofs. I understand why one of my symbolizations was incorrect, I used a biconditional when it should have been a conditional, and I understand a proof, I misused a rule, but as far as the others I got wrong I am quite confused. I am going to talk to my TA, but I wanted to see if I could get some other explanations as he can be somewhat difficult to understand.
SYMBOLIZATIONS
The abbreviations we were given to use are as follows :
C: Medical Salaries are curtailed
M: The medical system will be reformed
P: The poor will get quality medical care
R: The insurance companies are regulated
1.) The medical system will not be reformed and the poor will not get quality care, unless salaries are curtailed and insurance companies are regulated.

I answered (C&R) -> (M&P) and the expected answer reviewed in Discussion session today was -(C&R) -> (-M&-P), which I can now see as being the literal translation, but I feel like mine was intuitively equivalent. Am I way off base here? Looking at my paper it seemed like he was taking off 3 points for this out of 50, and this was similar to the second symbolization I don't get, he took the same amount off. Is this really worth over a letter grade of deduction?

2.) The medical system will not be reformed and the poor will not get quality care, unless medical salaries are curtailed or insurance companies are regulated.
Similar to above, my answer was, (CvR) -> (M&P) and the expected answer was -(CvR) -> (-M&-P).

PROOFS
As far as proofs go there is one I had a problem with.
We were given SvK, -KvG and told to prove GvS.
What follows is exactly what I wrote on the exam, so if there are unnecessary lines that is why.
1 1. SvK A
2 2. -KvG A
1 3. -S -> K 1AR
1 4. -K -> S 3 CN
2 5. K ->G 2 AR
6 6. -G PA
2 7. -G -> -K 5 CN
2,6 8. -K 6,7 ->O
2,3,6 9. S 8,4 -> O
2,3,6 10. GvS 9vI
^ 11. S -> (GvS) 9-10 ->I
12 12. K PA
2,12 13. G 5,12 -> O
2,12 14. GvS 13vI
2 15. K ->(GvS) 12-14 -> I
1,2 16. GvS 1,11,15 vO
My TA's notes were :
Line 11 - S Should be your PA if you use -> I
Line 12 - You need --K

I am not trying to have you do homework for me, in fact solutions were given during class, one such solution to the above proof was to use the Chain Argument rule, which I didn't see at the time. I just want other people to look at these and tell me what they think. I don't understand why I lost another letter grade worth of points on this proof... I think it follows?
 
kennethamy
 
Reply Wed 15 Apr, 2009 06:22 am
despinozist wrote:
"A unless B" is ~B > A

"unless" signifies an antecedent.



And, ~B >A is equivalent to, B v A. Which is equivalent to, ~(B & A). Thus the class answer, ~(C & R).
 
Nimmy
 
Reply Wed 15 Apr, 2009 06:49 am
@Nimmy,
First off, thanks to both of you for replying Smile

Despinozist - when you say that or arguments are ugly, trust me I agree with you! It was just the only way that clicked in my mind during the exam itself. My question was if it well any less valid than say... using chain argument once you've rearranged the premises a few times. One of the reasons the or argument came to my mind was that I had done a proof using it on a homework assignment and he had marked it as a "fun" way to derive SvT. The inconsistency was what really made me wonder if there was a large blunder in my moves.

and Kennethamy - That was my next question! Thanks!
 
kennethamy
 
Reply Wed 15 Apr, 2009 07:50 am
despinozist wrote:
I was only speaking of the behavior of "unless."

You don't want to take what kennethamy said at his letter. Since we're talking about "(C & R)", when I say "~A > B"

"(C & R)" just is "A".

So "~(C & R) > B".

I was explaining why the "(C & R)" has the outside negation.

Plus, B v A is not equivalent to ~(B & A)

It's equivalent to ~(~B & ~A). So kennethamy is wrong.


Right you are. Forgot the negation.
 
Nimmy
 
Reply Wed 15 Apr, 2009 09:48 am
despinozist;58265 wrote:
I was only speaking of the behavior of "unless."

You don't want to take what kennethamy said at his letter. Since we're talking about "(C & R)", when I say "~A > B"

"(C & R)" just is "A".

So "~(C & R) > B".

I was explaining why the "(C & R)" has the outside negation.

Plus, B v A is not equivalent to ~(B & A)

It's equivalent to ~(~B & ~A). So kennethamy is wrong.


I had to read this about four times, but it makes a ton of sense now! Laughing

Thanks! I am going to go talk to my T.A. soon, so maybe he can help me understand where my proof went haywire. I've really got to get all this down because we've just started Predicate Logic.

____________________________

Just in case anyone was wondering, the reason this proof was marked down was my execution of the Arrow In Rule. Smile
 
 

 
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