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Urgent Help Q2 Simplified...

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Horace phil
 
Reply Sun 12 Jul, 2009 06:34 pm
This is to simplify matters...

As I said in my other post, I am particularly grateful for the patience of VideCorSpoon and Goapy. I enjoy philosophy but am having trouble getting my head around some of these logical matters.

So what I need to do in this question is determine what binary truth-functional sentential connective stands in place of "#," in light of this information:

P#P is a tautology; and the sentence Q#P is not a tautological consequence of the sentences [P#Q] and P.

_________________________________________

It was my understanding that I was working with four possible connectives ("and", "or", "if/then" and "if and only if"). However (and I could be mistaken) I determined that neither of the "and" or "or" can be reconciled with P#P, while neither the "if/then" or "if and only if" can be reconciled with the formula I believed myself to be working with:

([P->Q] ^ P]) -> [Q-P]

It could be also however that I have interpretted the information wrong. Either way, I could use, and am grateful for, any confirmation or denial of my claims regarding the four common connectives...
 
kennethamy
 
Reply Sun 12 Jul, 2009 07:00 pm
@Horace phil,
Horace;76899 wrote:
This is to simplify matters...

As I said in my other post, I am particularly grateful for the patience of VideCorSpoon and Goapy. I enjoy philosophy but am having trouble getting my head around some of these logical matters.

So what I need to do in this question is determine what binary truth-functional sentential connective stands in place of "#," in light of this information:

P#P is a tautology; and the sentence Q#P is not a tautological consequence of the sentences [P#Q] and P.

_________________________________________

It was my understanding that I was working with four possible connectives ("and", "or", "if/then" and "if and only if"). However (and I could be mistaken) I determined that neither of the "and" or "or" can be reconciled with P#P, while neither the "if/then" or "if and only if" can be reconciled with the formula I believed myself to be working with:

([P->Q] ^ P]) -> [Q-P]

It could be also however that I have interpretted the information wrong. Either way, I could use, and am grateful for, any confirmation or denial of my claims regarding the four common connectives...


What does Q-P mean? What is the connective? Those are the four common connectives, though.
 
goapy
 
Reply Sun 12 Jul, 2009 07:18 pm
@kennethamy,
Horace, I answered this question in the other thread; here

Your original question:
Horace;76721 wrote:

2. I need to figure out what this binary truth-functional sentential connective (#) actually is in light of
P#P being a tautology, and Q#P not being a tautological consequence of the sentences P#Q and P.


If you replace all instances of the connective "#" with the connective of converse implication (as explained in the other post), then all of the statements in your question will be true.

I don't know if I can explain it any better than that. Carefully re-read my post in the other thread.
 
Horace phil
 
Reply Sun 12 Jul, 2009 07:19 pm
@Horace phil,
That's a typo.
That was meant to indicate "if/then" which I had tested out to see if it would fit, since P->P is tautological...
See the key is to find what the connective is.
Even if I replace that "->" with the "if and only if" it still doesn't seem to work...
 
goapy
 
Reply Sun 12 Jul, 2009 07:27 pm
@Horace phil,
Horace;76909 wrote:
That's a typo.
That was meant to indicate "if/then" which I had tested out to see if it would fit, since P->P is tautological...
See the key is to find what the connective is.
Even if I replace that "->" with the "if and only if" it still doesn't seem to work...


Horace, have you read my other post here?

Yes, both the "if/then" and the "if and only if" connectives will not work. I mentioned this three posts ago, and told you explicitly that none of the common four binary connectives will work. You must use one of the other sixteen binary connectives. I told you which one works. It seems that you're not picking up on this.
 
Horace phil
 
Reply Sun 12 Jul, 2009 07:37 pm
@Horace phil,
Yes I have read it. I had to double check though. I thought it odd that he would expect us to identify a connective that he had never informed us existed. Anyways, I feel close but this is the formula I have.

([Phttp://upload.wikimedia.org/math/a/6/4/a6465c0244621c63e7e1e96eb55aad7a.pngQ] and P]) http://upload.wikimedia.org/math/a/6/4/a6465c0244621c63e7e1e96eb55aad7a.png ([Q http://upload.wikimedia.org/math/a/6/4/a6465c0244621c63e7e1e96eb55aad7a.png P])

P http://upload.wikimedia.org/math/a/6/4/a6465c0244621c63e7e1e96eb55aad7a.pngQ means P v -Q
P
Q http://upload.wikimedia.org/math/a/6/4/a6465c0244621c63e7e1e96eb55aad7a.png P means Q v -P

So is this what we are dealing with:

([Phttp://upload.wikimedia.org/math/a/6/4/a6465c0244621c63e7e1e96eb55aad7a.pngQ] and P]) http://upload.wikimedia.org/math/a/6/4/a6465c0244621c63e7e1e96eb55aad7a.png ([Q http://upload.wikimedia.org/math/a/6/4/a6465c0244621c63e7e1e96eb55aad7a.png P])

I am putting it in a table at this moment...
 
goapy
 
Reply Sun 12 Jul, 2009 07:50 pm
@Horace phil,
Horace;76913 wrote:


So is this what we are dealing with:

([P←Q] and P]) ← ([Q ← P])

I am putting it in a table at this moment...


No, it is:

([P ← Q] & P) → (Q ← P)

Note: I changed the main connective of the sentence because the main connective is material implication. You are still checking if "Q ← P" is a consequence of "[P ← Q] and P"

It would be much easier to construct a truth table of the above if you replace it with this equivalent form:

([P v ~Q] & P) → (Q v ~P)
 
Horace phil
 
Reply Sun 12 Jul, 2009 07:54 pm
@Horace phil,
You deserve a medal. Smile
I'm just plugging this into a table.
At the beginning of today my goal was to figure out these questions, and while I haven't even remotely been able to do so independently, certain progress is being made.
Thanks again...

---------- Post added 07-12-2009 at 08:56 PM ----------

I have graphed it, and in the second line the connective -> bears a false, which makes this work I think. I will have to check over everything tomorrow...
 
 

 
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