View Full Version : Help in Logic
Bretty123
19th December 2006, 23:01
On Rosa's recommendation I picked up a used copy of E.J. Lemmon's Beginning Logic and although I have already learned alot of the beginning logic I have some questions.
My first question so far in reading it is: What is the point of the double negation rule?
Is it's application to just reverse the negation of a proposition?
Rosa Lichtenstein
19th December 2006, 23:54
It allows you to discharge propositions in simple/complex inferences, such as (on page 14, using '/' for the syntactic proof sign, ie., the letter 'T' on its side) this:
8 -P arrow Q, -Q/P
1 (1) -P arrow Q [A]
2 (2) -Q [A]
1,2 (3) --P [1,2 MTT]
1,2 (4) P [3 DN]
Without that rule this would not go through.
Recall, this sort of logic is about rules, not intuition. So, if you have a rule, you apply it rigorously.
If you want to question that rule, that is another story.
bretty
30th December 2006, 15:00
Rosa I'm having trouble with these three while doing the problems. I may have just been burnt out or overlooked something but I couldn't figure them out.
P*R, -Q*R, PvQ therefore R
and
(PvQ) <-> P therefore Q -> P
and
P * P therefore P
For the last one do you just assume P and derive a contradiction via RAA?
Rosa Lichtenstein
30th December 2006, 15:45
Bretty, you will need to make this clearer if I am to help you.
Your symbol '*' is undefined, for starters.
And you do not distinguish premisses from inferences.
If you are appealing to something in Lemmon, it will be easier if you just append a page or exercise reference.
bretty
9th January 2007, 20:44
Rosa forget about those other questions but I have another one regarding page 41 question C.
Also I'm having trouble figuring out in what context I need to use the rule Velimination?
Could you please help me out? I love logic but I'm having trouble figuring out these rules sometimes especially without an answer book for this.
Also I'm taking Logic at university now however we are using Donald Kalish' book of logic which is used heavily apparently around my area in academics.
And to add one more can you show me how to do this question? I cannot for the life of me figure it out.
(PvQ) & (PvR) therefore P v (Q & R)
Thanks a bunch.
Rosa Lichtenstein
10th January 2007, 02:26
Everyone finds this rule problematic!
It is perhaps easier to see this rule if we translate it into ordinary English; vE is the equivalent of 'either way'. And, it is what is called a 'thematic' rule, as opposed to a schematic' rule (that is, it is a way of knitting arguments (not atomic propositions) together that are already valid); and if we use a simpler argument to get this idea across.
So, we can argue thus (but 'conditionalising’ this at first, just to get the idea across -- i.e., using 'if...then'):
If the US leaves Iraq it is screwed.
On the other hand, if they do not, they are screwed.
So either way, the US is screwed.
Let P = "The US quits Iraq"; R = "The US is screwed"
And Q = "The US stays in Iraq", but now using "/" as the syntactic argument sign (i.e., as 'therefore').
1) P/R. [I.e., 'P therefore R'. Assuming a valid argument can be found that enables R to be derived from P!]
2 Q/R. [Ditto]
3 So PvQ/R.
More colloquially:
1) The US says in Iraq therefore the US is screwed (assuming this can be shown to be valid by some devious means).
2) The US quits Iraq therefore the US is screwed (ditto).
3) Either way, the US is screwed.
Now this rule applies in a more complex way in the example you refer to. And the one on page 41 is inter-derivable:
I.e., P v (Q & R) / (P v Q) & (P v R); and:
(P v Q) & (P v R) / P v (Q & R);
I.e., using % for inter-derivability: P v (Q & R) % (P v Q) & (P v R).
However, you just want a proof of the latter; here goes (but it's years since I have done this sort of thing, so the numbering might need to be checked!!).
The formatting here has eliminated the gaps I put in to make the line numbers and rules used clearer, so I have now put them in bold so they stand out better:
(P v Q) & (P v R) / P v (Q & R).
1 (1) (P v Q) & (P v R) A
1 (2) (P v Q) 1&E
1 (3) (P v R) 1&E
4 (4) Q A
5 (5) R A
1,4,5 (6) Q & R 1,2,3,4,5 &I
1,4,5 (7) P v (Q & R) 1,2,3,4,5,6 vI
8 (8) P A
1,8 (9) P v (Q & R) 1,2,3,8 vI
1,4,5,8 (10) P v (Q & R) 1,1,2,2,3,3,4,5,6,7,8,9vE
1 (11) P v (Q & R) 1,2,3,8,10vE
Recall that in this area of logic you can assume anything you like (but naturally, one will only assume the relevant letters).
So, if I can show that I can arrive at P v (Q & R) either way (by the route from 2 to 7, or from 8 to 9), then the rule of vE allows me to derive the given result at line 10 ('either way').
The last line is needed to show that the required result follows from the initial assumption alone.
Can I suggest you obtain George Schumm's 'A Teaching Companion to Lemmon's Beginning Logic' (Hackett Publishing 1979)??
http://www.amazon.co.uk/gp/search?search-a...bmitbutton1.y=8 (http://www.amazon.co.uk/gp/search?search-alias=stripbooks&field-keywords=&author=Schumm%2C+G&select-author=field-author-like&title=%27A+Teaching+Companion+to+Lemmon%27s+Beginn ing+Logic&select-title=field-title&subject=&select-subject=field-subject&field-publisher=&field-isbn=&chooser-sort=rank%21%2Bsalesrank&node=&field-binding=&mysubmitbutton1.x=36&mysubmitbutton1.y=8)
At the UK Amazon it is going for £2.24, i.e., about $4.00.
[I have just checked; there is a solution to this question on page 51, which is almost identical to mine; but the numbering is different, and the author proves my second half first -- which is neither here nor there.]
It will help you.
Hope this helps too!
Rosa Lichtenstein
10th January 2007, 02:26
Double Post!
bretty
10th January 2007, 05:07
I dont understand why you have so many numbers to the right of the proofs. In the book it shows just the ones that the proof involves. Where do all those lines of numbers come from? Its hard for me to tell which ones are the 5 you necessarily have to write down for V-elim.
Thanks again this was helpful!
Rosa Lichtenstein
10th January 2007, 09:57
Bretty, as I told you, the numbers are probably incorrect, since it is years since I used to do such basic proofs.
Put your own numbers in; they will probably be more correct than mine.
But the proof itself is OK.
bretty
10th January 2007, 15:03
Okay thank you very much. Your pretty good at this. I had your exact proof but I couldn't figure out how to use V-Elim. properly.
Rosa Lichtenstein
10th January 2007, 15:47
Well done for sticking at it; the important thing with logic is not to give up -- unless one has a flair for it (Lemmon certainly did; he apparently learnt the whole of modern logic in 3 months!), it does take time.
I have studied advanced group theory in algebra, and compared to logic (and especially the philosophjy of logic), it was a doddle.
Logic and the Philosophy of Logic are the most dufficult subjects the human mind has yet invented, in my opinion.
Rosa Lichtenstein
10th January 2007, 15:55
Bretty, check out Natural Deduction on the web, for instance here:
http://www.danielclemente.com/logica/dn.en.html
http://www.swif.uniba.it/lei/foldop/foldoc...tural+deduction (http://www.swif.uniba.it/lei/foldop/foldoc.cgi?natural+deduction)
http://en.wikipedia.org/wiki/Natural_deduction
bretty
20th January 2007, 19:18
Okay Rosa I'm taking Logic at university right now and they are making us use a different way of constructing derivations. We are using Donald Kalish and Montague's book Logic: Techniques of formal reasoning and I'm completely confused on how to use this new method.
Have you heard of it?
Basically we have to use the Show method where an example would be
(Q then S) then T, T then Q, therefore Q
Show Q etc.
And then when we are finished we have to box things in.. have you seen this method before?
Rosa Lichtenstein
20th January 2007, 20:29
Yes, I have heard of it, but I have not studied it.
There are 1001 different forms of logic these days.
Good luck with it! I do not think I will be able to help you much.
Comrade Red might be able to, though.
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