19 May 12
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.
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19 May 12
Originally posted by LemonJelloIs 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.)
🙂
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.
19 May 12
2 edits
Originally posted by Conrau KCan 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.
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.)
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.
19 May 12
Originally posted by sonhouseYou are right. However, as I said before, it amounts to nonsense because of the error of using P and Q. 😀
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.
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19 May 12
Originally posted by sonhouseSure 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'.
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.
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.
20 May 12
1 edit
Originally posted by Conrau KHere is I guess what used to be a preprint of the paper:
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.)
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.
20 May 12
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?
20 May 12
1 edit
Originally posted by sonhouseAs 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):
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?
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.