Proof that God Exists

Originally posted by Conrau K
[b] The problem here must lie in how it treats an indicative conditional like 'If I am evil, I will be punished after I die' and bases it in the truth functionality of a material conditional.

Yes, that sums it up. It was nothing to do with modus tollens per se. Thank you for that interesting piece of logic.[/b]

🙂

I found it rather interesting too. I was reading a paper on indicative conditionals by Barbara Abbott and she introduced this exercise. She attributed the idea to Michael Jubien.

Originally posted by LemonJello
Yes, very good. You've proven yourself much more worthy than our friend Zahlanzi.

I don't know. Zahlanzi seems to be holding on to 4th place on this thread so far. Could have been worse if more people had posted.

Thank you LJ and Conrau.

Originally posted by LemonJello
Tell me precisely what is wrong with the proof. I know what the problem is here; do you?

What is wrong with the proof is that you are using P and Q. 😀

Originally posted by LemonJello
🙂

I found it rather interesting too. I was reading a paper on indicative conditionals by Barbara Abbott and she introduced this exercise. She attributed the idea to Michael Jubien.

Is this paper available on Jstore? I am interested in how you could develop a logic to account for the invalidity of that argument. Do you just rule out embedded conditionals (which seems likely to be a harsh thing to do) or is there a non-classical logic she recommends? (I haven't done logic in years but am enthused hobbyist on the subject.)

Originally posted by Conrau K
Is this paper available on Jstore? I am interested in how you could develop a logic to account for the invalidity of that argument. Do you just rule out embedded conditionals (which seems likely to be a harsh thing to do) or is there a non-classical logic she recommends? (I haven't done logic in years but am enthused hobbyist on the subject.)

Can you explain the symbols you used in your logic exercise, what exactly they mean? I haven't studied logic and don't know what they mean. I think the ~ sign means 'not'? That's about as far as I got.

P -> Q
Not-Q.
Therefore, not-P.

I thought in math the > symbol meant the symbol on the left is greater than the symbol directly to the right of it but it seems to mean something different here.

Originally posted by sonhouse
Can you explain the symbols you used in your logic exercise, what exactly they mean? I haven't studied logic and don't know what they mean. I think the ~ sign means 'not'? That's about as far as I got.

P -> Q
Not-Q.
Therefore, not-P.

I thought in math the > symbol meant the symbol on the left is greater than the symbol directly to the right of it but it seems to mean something different here.

You are right. However, as I said before, it amounts to nonsense because of the error of using P and Q. 😀

Originally posted by RJHinds
You are right. However, as I said before, it amounts to nonsense because of the error of using P and Q. 😀

So he is not minding his P's and Q's? Would it have been a better argument if he had used A's and B's?

Originally posted by sonhouse
Can you explain the symbols you used in your logic exercise, what exactly they mean? I haven't studied logic and don't know what they mean. I think the ~ sign means 'not'? That's about as far as I got.

P -> Q
Not-Q.
Therefore, not-P.

I thought in math the > symbol meant the symbol on the left is greater than the symbol directly to the right of it but it seems to mean something different here.

Sure thing. P, Q and R, are propositional variables. They stand for propositions ('God exists' 'I will punished in Hell' 'I am evil'😉 but it is not necessarily important to know the propositional content of these variables is. ~ is an operator which you rightly note means 'not'. --> is a conjunction meaning 'if'. The syntax works like this, P-->Q means 'If P, then Q'; ~P-->Q means 'If not-P, then Q'; ~(P-->Q) means 'It is not the case that if P, then Q'.

It is important to understand for this exercise that a conditional P-->Q is true in either two ways: if P is false or Q is true. This may sound strange because we think of a conditional as being true only if Q follows as a logical consequence of P. But think of it this way, a conditional is only false if the antecedent, P, is true and the consequent is false. So a truth table is like this where 0=false, 1=true.

P Q P-->Q
1 1 1
1 0 0
0 1 1
0 0 1

Only in one of four cases is the conditional P-->Q false.

Now in LemonJello's case the logic works like this:
(1) ~P --> ~(Q-->R)
(2) ~Q
(3) P

~Q, therefore Q-->R (if Q is false, then the conditional is trivially true, as in the table above).

So ~(Q-->P) is false.

But from (1) if ~(Q-->P) is false, then the conditional could only still be true if ~P is also false. So ~~P. hence P, God exists.

Clearly this is a failure of classical logic and we would basically want to reject triviality like ~Q, therefore Q-->R.

Originally posted by Conrau K
Is this paper available on Jstore? I am interested in how you could develop a logic to account for the invalidity of that argument. Do you just rule out embedded conditionals (which seems likely to be a harsh thing to do) or is there a non-classical logic she recommends? (I haven't done logic in years but am enthused hobbyist on the subject.)

Here is I guess what used to be a preprint of the paper:

https://www.msu.edu/~abbottb/indconds.pdf

In the paper, unfortunately, she does not purport to give any real resolution for how indicative conditionals ought to be treated (she tends to think we ought to stick to something like a material conditional approach, but even with modifications she still sees several lingering difficulties with that approach). But the paper tries to highlight some problems with various approaches, which I found interesting.

Originally posted by sonhouse
So he is not minding his P's and Q's? Would it have been a better argument if he had used A's and B's?

You are a wiser man than I thought.

http://en.wikipedia.org/wiki/Mind_your_Ps_and_Qs

Originally posted by RJHinds
You are a wiser man than I thought.

http://en.wikipedia.org/wiki/Mind_your_Ps_and_Qs

Interesting but not exactly relevant to the op.

It is important to understand for this exercise that a conditional P-->Q is true in either two ways: if P is false or Q is true. This may sound strange because we think of a conditional as being true only if Q follows as a logical consequence of P. But think of it this way, a conditional is only false if the antecedent, P, is true and the consequent is false. So a truth table is like this where 0=false, 1=true.

P Q P-->Q
1 1 1
1 0 0
0 1 1
0 0 1

Only in one of four cases is the conditional P-->Q false.

Now in LemonJello's case the logic works like this:
(1) ~P --> ~(Q-->R)
(2) ~Q
(3) P

I seem to get the second part, but which one of the four cases is the conditional false?

Originally posted by sonhouse
It is important to understand for this exercise that a conditional P-->Q is true in either two ways: if P is false or Q is true. This may sound strange because we think of a conditional as being true only if Q follows as a logical consequence of P. But think of it this way, a conditional is only false if the antecedent, P, is true and the consequent is fals P

I seem to get the second part, but which one of the four cases is the conditional false?

As Conrau mentioned, the material conditional P-->Q is false only in the case where the antecedent P is true (=1) and the consequent Q is false (=0):

P Q P-->Q
1 1 1
1 0 0 <--------This case is the only one where (P-->Q) is false (=0).
0 1 1
0 0 1

The reason why I think the "proof" in the OP is interesting is because it is logically valid under a (commonly assumed) logic in which we treat the conditionals therein as having the truth functionality of the material conditional outlined above. But this result seems unsatisfactory. So, I think it shows we need a modified or different treatment of such conditionals. The paper to which I provided a link has some discussion on this.