Propositional Logic (Part 2)

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Re: Propositional Logic (Part 2)

Postby someguy1 on September 8th, 2017, 4:00 am 

BadgerJelly » September 8th, 2017, 1:55 am wrote:Mitch said -

The truthteller's answers all have the truth-value T, so if left is correct, he would answer (in reply to the question "Is left correct?"): "Yes, left is correct". So, for the truthteller, the truthful answer to Q is Yes; and he truthfully answers it "Yes".

The liar's answers all have the truth-value F, so if left is correct, he would answer (in reply to the question "Is left correct?"): "No, left is not correct". So, for the liar, the truthful answer to Q is No; but (being a liar) he falsely answers it "Yes".

If right is correct, then for the truthteller the truthful answer to Q is No; and he truthfully answers it "No".

If right is correct, then for the liar the truthful answer to Q is Yes (he would say left is correct in order to lie consistently); but he falsely answers Q "No".


I completely agree. This is what I meant by a "double-negative" in the answer. The correct, and truthful, answer is squeezed out of the liar.

Someguy -

Do you really see a problem with the above? I think the main confusion arises due to mixing up "truth value" with "truth".


I agree with what you quoted. But that's only the warmup to the question. The statement S is false because it's not a biconditional. But I've said that quite a number of times (without anyone logically refuting me) and my saying it more isn't going to help.

But ok, one last time for the night.

Two statements: P = "The correct door is on the left"; and Q = "The speaker is a truthteller."

We wish to evaluate the truth value of P <=> Q or P iff Q (these mean the same thing).

To do this, we have to show that P => Q and Q => P. You can't do that and Mitch can't either, because neither implication is true.

So the biconditional P <=> Q is false. Therefore the truthteller will ALWAYS say it's false and the liar will ALWAYS say it's true, in either case giving no information about the door.

And like I say, if I'm wrong, somebody show me I'm wrong. I'm open to the possiblity. To show me I'm wrong you have to show that P <=> Q is true. How can you do that? I claim you can't. Proof: What if the good door is on the left and the speaker is a liar? Then P and Q have opposite truth values. QED. I'm going to bed.
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Re: Propositional Logic (Part 2)

Postby BadgerJelly on September 8th, 2017, 4:13 am 

Someguy -

I have to admit you're making very little sense. You agree with it, but dispute it at the same time?

Is this a biconditional statement:

"All bananas are square if and only if they are green."
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Re: Propositional Logic (Part 2)

Postby Positor on September 8th, 2017, 8:49 am 

BadgerJelly » September 8th, 2017, 1:55 am wrote:Mitch said -
The truthteller's answers all have the truth-value T, so if left is correct, he would answer (in reply to the question "Is left correct?"): "Yes, left is correct". So, for the truthteller, the truthful answer to Q is Yes; and he truthfully answers it "Yes".

The liar's answers all have the truth-value F, so if left is correct, he would answer (in reply to the question "Is left correct?"): "No, left is not correct". So, for the liar, the truthful answer to Q is No; but (being a liar) he falsely answers it "Yes".

If right is correct, then for the truthteller the truthful answer to Q is No; and he truthfully answers it "No".

If right is correct, then for the liar the truthful answer to Q is Yes (he would say left is correct in order to lie consistently); but he falsely answers Q "No".

Actually, the embedded quote above is by me, not Mitch. :)

someguy1 wrote:Two statements: P = "The correct door is on the left"; and Q = "The speaker is a truthteller."

This is not quite right. The two statements we are actually concerned with in this problem are:

P = "The correct door is on the left."
Q = "The speaker will say the correct door is on the left iff he is the truthteller."

I think you will agree that P and Q now make a biconditional.
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Re: Propositional Logic (Part 2)

Postby mitchellmckain on September 8th, 2017, 2:02 pm 

This thread has left the science of symbolic logic behind to indulge in the empty rhetoric of personal attacks and incorrect quotations added to incorrect understanding of symbolic logic contribute nothing but confusion to the topic. I am no longer interested and will no longer waste my time with it.
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Re: Propositional Logic (Part 2)

Postby BadgerJelly on September 18th, 2017, 3:42 am 

Statements

Statement p, "is it the case the door is the correct door"

Statement q, "you are a truth teller"

biconditional (p<=>q)

Is the same as

(p=>q) & (q=>p)

So we have two statements in the biconditional

1) IF you are a truth teller THEN is it the case that the door is the correct door?

2) IF the door is the correct door THEN is it the case that you are a truth teller?

I am having real trouble dealing with "statements" as "questions".

Mitch -

It would've helped me more if you had stated what I was saying correctly and what I was saying incorrectly. It is my thread not someguy's. I took some liberty by saying "double negative" previously.

I am trying to understand the problem intuitively.

note: I would add that I am coming across quite a few different notations that don't add up. Some people use => where others use ->, and other instances this symbol <-> is used as biconditional and this <=> only used for equivalence.
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Re: Propositional Logic (Part 2)

Postby mitchellmckain on September 18th, 2017, 4:37 am 

BadgerJelly » September 18th, 2017, 2:42 am wrote:It is my thread not someguy's.

Good point! I guess I just excuse myself from engaging anyone using such tactics.

BadgerJelly » September 18th, 2017, 2:42 am wrote:Statements

Statement p, "is it the case the door is the correct door"

Statement q, "you are a truth teller"

biconditional (p<=>q)

Is the same as

(p=>q) & (q=>p)

So we have two statements in the biconditional

1) IF you are a truth teller THEN is it the case that the door is the correct door?

2) IF the door is the correct door THEN is it the case that you are a truth teller?

I am having real trouble dealing with "statements" as "questions".

Interesting approach... maybe?

But the correct way to do this is to keep the conjunction... So the original question is equivalent to ...
1. If you are a truth teller, then the door is the correct one.
2. If the door is the correct one, then you are a truth teller.
Are both 1 and 2 correct?

First the easy part. Clearly the truth teller will say yes only if the door is the correct one and say no if the door is not the correct one.

As for the liar, if he says yes then we know that either 1 or 2 is false (or both). Since he is not the truth teller then 1 is automatically true. So we know that 2 has to be false. But again he is not the truth teller, and the implication a->b is not false when both a and b are false, but only when a is true but b is false. Thus the door has to be the correct one.

If the liar says no, then we know that both 1 and 2 are true. But since he is not the truth teller then once again 1 is automatically true. But for 2 to be true, the door cannot be the correct one.

Thus, whether the person answering the question is truth teller or liar, it will be the answer to whether the door is the correct one.

in the end, it works, but I am not sure this has helped to make it intuitive... oh well

But another identity you can use is...

(not p and not q) or (p and q)

With this the question becomes...
1. This is not the correct door and you are not a truth teller.
2. This is the correct door and you are a truth teller.
Is either 1 or 2 true?

Perhaps this is more intuitive because you can say that for the truth teller only 2 can be true and only if it is the correct door, while for the liar only 1 can be true and only if it is not the correct door. But as a liar, that means he will only say no if it is not the correct door. Thus once again both truth teller and liar will answer yes if it is the correct door and no if it is not the correct door.

does that help?
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Re: Propositional Logic (Part 2)

Postby mitchellmckain on September 19th, 2017, 12:02 am 

In other words

If it the truth teller then p is true and in that case
(not p and not q) or (p and q) = q
If it is the liar then p is false and in that case
(not p and not q) or (p and q) = not q

Thus (not p and not q) or (p and q) is effectively choosing the answer the person gives based on whether he is the truth teller or the liar. Make sense?


As for where this identity came from....
p<=>q == (not p and not q) or (p and q)
You can google a list of identities for symbolic logic or...
you can practically read the rhs off the truth table for p<=>q.
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Re: Propositional Logic (Part 2)

Postby BadgerJelly on September 19th, 2017, 1:26 am 

Thanks for trying Mitch. I've found a good set of vids on youtube that I am working through. They have been much more helpful than the stuff on the courser course.

I am a stubborn soab so I'll find a way to consolidate the linguistics and logic eventually.

For anyone interesting in pursuing what I am I recommend looking at this site (pretty much covers all the basics of what I am doing this for and he obviously has similar interests to me):

https://www.youtube.com/user/thetrevtutor/playlists
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Re: Propositional Logic (Part 2)

Postby BadgerJelly on September 19th, 2017, 3:04 am 

And MITCH ... My big problem is with the QUESTION. How does the "?" fit into Propositional Logic? This is my number problem, as you can see in the EQUIVALENCE of the following:

(p->q) & (q->p) <=> (p<->q)

In the question presented in the OP we have the form (p<->q) ?

How does this "?" present itself for "(p->q) & (q->p)" in such a way to make it equivalent to "(p<->q)?" ?

I think if I understood that I'd be able to think of this more intuitively.

Note: I can find plenty of examples online but none that deal with the statement posed as a question.
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Re: Propositional Logic (Part 2)

Postby mitchellmckain on September 19th, 2017, 7:11 pm 

BadgerJelly » September 19th, 2017, 2:04 am wrote:And MITCH ... My big problem is with the QUESTION. How does the "?" fit into Propositional Logic? This is my number problem, as you can see in the EQUIVALENCE of the following:

(p->q) & (q->p) <=> (p<->q)

In the question presented in the OP we have the form (p<->q) ?

How does this "?" present itself for "(p->q) & (q->p)" in such a way to make it equivalent to "(p<->q)?" ?

I think if I understood that I'd be able to think of this more intuitively.

Note: I can find plenty of examples online but none that deal with the statement posed as a question.


As I made clear above, you cannot distribute the question mark, especially when dealing with liars and truth tellers. The liar only lies to the overall question. I won't say that distributing the question mark is impossible but I will say that it would be inventing something outside of symbolic logic.
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