Sorry, I think I still don’t understand your reasoning.
First, I have the beliefs P1, P2 and P3, then I (in an apparently deductively valid way) reason that [C1] “T is a theorem of P1, P2, and P3”, therefore I believe T.
Either my reasoning that finds out [C1] is valid or invalid. I do think it’s valid, but I am fallible.
Then the Authority asserts F, I add F to the belief pool, and we (in an apparently deductively valid way) reason [C2] “~T is a theorem of F, P1, P2, and P3”, therefore we believe ~T.
Either our reasoning that finds out [C2] is valid or invalid. We do think it’s valid, but we are fallible.
Is it possible to conclude C2 without accepting I made a mistake when reasoning C1 (therefore we were wrong to think that line of reasoning was valid)? Otherwise we would have both T and ~T as theorems of F, P1, P2, and P3, and we should conclude that the promises lead to contradiction and should be revised; we wouldn’t jump from believing T to believing ~T.
But the story doesn’t say the Authority showed a mistake in C1. It says only that she made a (apparently valid) reasoning using F in addition to P1, P2, and P3.
If the argument of the Authority doesn’t show the mistake in C1, how should I decide whether to believe C1 has a mistake, C2 has a mistake, or the promises F, P1, P2, and P3 actually lead to contradiction, with both C1 and C2 being valid?
I think Bayesian reasoning would inevitably enter the game in that last step.
C1 is a presumption, namely, a belief in the truth of T, which is apparently a theorem of P1, P2, and P3. As a belief, it’s validity is not what is at issue here, because we are concerned with the truth of T.
F comes in, but is improperly treated as a premiss to conclude ~T, when it is equivalent to ~T. Again, we should not be concerned with belief, because we are dealing with statements that are either true or false. Either but not both (T or ~T) can be true (which is the definition of a logical tautology).
Hence C2 is another presumption with which we should not concern ourselves. Belief has no influence on the outcome of T or ~T.
For the first bullet: no, it is not possible, in any case, to conclude C2, for not to agree that one made a mistake (i.e., reasoned invalidly to T) is to deny the truth of ~T which was shown by Ms. Math to be true (a valid deduction).
Second bullet: in the case of a theorem, to show the falsity of a conclusion (of a theorem) is to show that it is invalid. To say there is a mistake is a straightforward corollary of the nature of deductive inference that an invalid motion was committed.
Third bullet: I assume that the problem is stated in general terms, for had Ms. Math shown that T is false in explicit terms (contained in F), then the proper form of ~T would be: F → ~T. Note that it is wrong to frame it the following way: F, P1, P2, and P3 → ~T. It is wrong because F states ~T. There is no “decision” to be made here! Bayesian reasoning in this instance (if not many others) is a misapplication and obfuscation of the original problem from a poor grasp of the nature of deduction.
(N.B.: However, if the nature of the problem were to consist in merely being told by some authority a contradiction to what one supposes to be true, then there is no logically necessity for us to suddenly switch camps and begin to believe in the contradiction over one’s prior conviction. Appeal to Authority is a logical fallacy, and if one supposes Bayesian reasoning is a help there, then there is much for that person to learn of the nature of deduction proper.)
Let me give you an example of what I really mean:
Note statements P, Q, and Z:
(P) Something equals something and something else equals that same something such that both equal each other.
(Q) This something equals that. This other something also equals that.
(Z) The aforementioned somethings equal each other.
It is clear that Z follows from P and Q, no? In effect, you’re forced to accept it, correct? Is there any “belief” involved in this setting? Decidedly not. However, let’s suppose we meet up with someone who disagrees and states: “I accept the truths of P and Q but not Z.”
Then we’ll add the following to help this poor fellow:
(R) If P and Q are true, then Z must be true.
They may respond: “I accept P, Q, and R as true, but not Z.”
And so on ad infinitum. What went wrong here? They failed to reason deductively. We might very well be in the same situation with T, where
(P and Q) are equivalent to (P1, P2, and P3) (namely, all of these premisses are true), such that whatever Z is, it must be equivalent to the theorem (which would in this case be ~T, if Ms. Math is doing her job and not merely deigning to inform the peons at the foot of her ivory tower).
P1, P2, and P3 are axiomatic statements. And their particular relationship indicates (the theorem) S, at least to the one who drew the conclusion.
If a Ms. Math comes to show the invalidity of T (by F), such that ~T is valid (such that S = ~T), then that immediately shows that the claim of T (~S) was false. There is no need for belief here; ~T (or S) is true, and our fellow can continue in the vain belief that he wasn’t defeated, but that would be absolutely illogical; therefore, our fellow must accept the truth of ~T and admit defeat, or else he’ll have departed from the sphere of logic completely.
Note that if Ms. Math merely says “T is false” (F) such that F is really ~T, then the form [F, P1, P2, and P3] implies ~T is really a circular argument, for the conclusion is already assumed within the premisses. But, as I said, I was being charitable with the puzzles and not assuming that that was being communicated.
I guess it wasn’t clear, C1 and C2 reffered to the reasonings as well as the conclusions they reached. You say belief is of no importance here, but I don’t see how you can talk about “defeat” if you’re not talking about justified believing.
For the first bullet: no, it is not possible, in any case, to conclude C2, for not to agree that one made a mistake (i.e., reasoned invalidly to T) is to deny the truth of ~T which was shown by Ms. Math to be true (a valid deduction).
I’m not sure if I understood what you said here. You agree with what I said in the first bullet or not?
Second bullet: in the case of a theorem, to show the falsity of a conclusion (of a theorem) is to show that it is invalid. To say there is a mistake is a straightforward corollary of the nature of deductive inference that an invalid motion was committed.
Are you sure that’s correct? If there’s a contradiction within the set of axioms, you could find T and ~T following valid deductions, couldn’t you? Proving ~T and proving that the reasoning leading to T was invalid are only equivalent if you assume the axioms are not contradictory. Am I wrong?
P1, P2, and P3 are axiomatic statements. And their particular relationship indicates (the theorem) S, at least to the one who drew the conclusion. If a Ms. Math comes to show the invalidity of T (by F), such that ~T is valid (such that S = ~T), then that immediately shows that the claim of T (~S) was false. There is no need for belief here; ~T (or S) is true, and our fellow can continue in the vain belief that he wasn’t defeated, but that would be absolutely illogical; therefore, our fellow must accept the truth of ~T and admit defeat, or else he’ll have departed from the sphere of logic completely.
The problem I see here is: it seems like you are assuming that the proof of ~T shows clearly the problem (i.e. the invalid reasoning step) with the proof of T I previously reasoned. If it doesn’t, all the information I have is that both T and ~T are derived apparently validly from the axioms F, P1, P2, and P3. I don’t see why logic would force me to accept ~T instead of believing there’s a mistake I can’t see in the proof Ms. Math showed me, or, more plausibly, to conclude that the axioms are contradictory.
...I don’t see how you can talk about “defeat” if you’re not talking about justified believing
“Defeat” would solely consist in the recognition of admitting to ~T instead of T. Not a matter of belief per se.
You agree with what I said in the first bullet or not?
No, I don’t.
The problem I see here is: it seems like you are assuming that the proof of ~T shows clearly the problem (i.e. the invalid reasoning step) with the proof of T I previously reasoned. If it doesn’t, all the information I have is that both T and ~T are derived apparently validly from the axioms F, P1, P2, and P3.
T cannot be derived from [P1, P2, and P3], but ~T can on account of F serving as a corrective that invalidates T. The only assumptions I’ve made are 1) Ms. Math is not an ivory tower authoritarian and 2) that she wouldn’t be so illogical as to assert a circular argument where F would merely be a premiss, instead of being equivalent to the proper (valid) conclusion ~T.
Anyway, I suppose there’s no more to be said about this, but you can ask for further clarification if you want.
2) that she wouldn’t be so illogical as to assert a circular argument where F would merely be a premiss, instead of being equivalent to the proper (valid) conclusion ~T.
Oh, now I see what you mean. I interpreted F as a new promiss, a new axiom, not a whole argument about the (mistaken) reasoning that proved T. For example, (wikipedia tells me that) the axiom of determinacy is inconsistent with the axiom of choice. If I had proved T in ZFC, and Ms. Math asserted the Axiom of Determinacy and proved ~T in ZFC+AD, and I didn’t know beforehand that AD is inconsistent with AC, I would still need to find out what was the problem.
I still think this is more consistent with the text of the original post, but now I understand what you meant by ” I was being charitable with the puzzles”.
Sorry, I think I still don’t understand your reasoning.
First, I have the beliefs P1, P2 and P3, then I (in an apparently deductively valid way) reason that [C1] “T is a theorem of P1, P2, and P3”, therefore I believe T.
Either my reasoning that finds out [C1] is valid or invalid. I do think it’s valid, but I am fallible.
Then the Authority asserts F, I add F to the belief pool, and we (in an apparently deductively valid way) reason [C2] “~T is a theorem of F, P1, P2, and P3”, therefore we believe ~T.
Either our reasoning that finds out [C2] is valid or invalid. We do think it’s valid, but we are fallible.
Is it possible to conclude C2 without accepting I made a mistake when reasoning C1 (therefore we were wrong to think that line of reasoning was valid)? Otherwise we would have both T and ~T as theorems of F, P1, P2, and P3, and we should conclude that the promises lead to contradiction and should be revised; we wouldn’t jump from believing T to believing ~T.
But the story doesn’t say the Authority showed a mistake in C1. It says only that she made a (apparently valid) reasoning using F in addition to P1, P2, and P3.
If the argument of the Authority doesn’t show the mistake in C1, how should I decide whether to believe C1 has a mistake, C2 has a mistake, or the promises F, P1, P2, and P3 actually lead to contradiction, with both C1 and C2 being valid?
I think Bayesian reasoning would inevitably enter the game in that last step.
C1 is a presumption, namely, a belief in the truth of T, which is apparently a theorem of P1, P2, and P3. As a belief, it’s validity is not what is at issue here, because we are concerned with the truth of T.
F comes in, but is improperly treated as a premiss to conclude ~T, when it is equivalent to ~T. Again, we should not be concerned with belief, because we are dealing with statements that are either true or false. Either but not both (T or ~T) can be true (which is the definition of a logical tautology).
Hence C2 is another presumption with which we should not concern ourselves. Belief has no influence on the outcome of T or ~T.
For the first bullet: no, it is not possible, in any case, to conclude C2, for not to agree that one made a mistake (i.e., reasoned invalidly to T) is to deny the truth of ~T which was shown by Ms. Math to be true (a valid deduction).
Second bullet: in the case of a theorem, to show the falsity of a conclusion (of a theorem) is to show that it is invalid. To say there is a mistake is a straightforward corollary of the nature of deductive inference that an invalid motion was committed.
Third bullet: I assume that the problem is stated in general terms, for had Ms. Math shown that T is false in explicit terms (contained in F), then the proper form of ~T would be: F → ~T. Note that it is wrong to frame it the following way: F, P1, P2, and P3 → ~T. It is wrong because F states ~T. There is no “decision” to be made here! Bayesian reasoning in this instance (if not many others) is a misapplication and obfuscation of the original problem from a poor grasp of the nature of deduction.
(N.B.: However, if the nature of the problem were to consist in merely being told by some authority a contradiction to what one supposes to be true, then there is no logically necessity for us to suddenly switch camps and begin to believe in the contradiction over one’s prior conviction. Appeal to Authority is a logical fallacy, and if one supposes Bayesian reasoning is a help there, then there is much for that person to learn of the nature of deduction proper.)
Let me give you an example of what I really mean:
Note statements P, Q, and Z:
(P) Something equals something and something else equals that same something such that both equal each other. (Q) This something equals that. This other something also equals that. (Z) The aforementioned somethings equal each other.
It is clear that Z follows from P and Q, no? In effect, you’re forced to accept it, correct? Is there any “belief” involved in this setting? Decidedly not. However, let’s suppose we meet up with someone who disagrees and states: “I accept the truths of P and Q but not Z.”
Then we’ll add the following to help this poor fellow:
(R) If P and Q are true, then Z must be true.
They may respond: “I accept P, Q, and R as true, but not Z.”
And so on ad infinitum. What went wrong here? They failed to reason deductively. We might very well be in the same situation with T, where
(P and Q) are equivalent to (P1, P2, and P3) (namely, all of these premisses are true), such that whatever Z is, it must be equivalent to the theorem (which would in this case be ~T, if Ms. Math is doing her job and not merely deigning to inform the peons at the foot of her ivory tower).
P1, P2, and P3 are axiomatic statements. And their particular relationship indicates (the theorem) S, at least to the one who drew the conclusion. If a Ms. Math comes to show the invalidity of T (by F), such that ~T is valid (such that S = ~T), then that immediately shows that the claim of T (~S) was false. There is no need for belief here; ~T (or S) is true, and our fellow can continue in the vain belief that he wasn’t defeated, but that would be absolutely illogical; therefore, our fellow must accept the truth of ~T and admit defeat, or else he’ll have departed from the sphere of logic completely. Note that if Ms. Math merely says “T is false” (F) such that F is really ~T, then the form [F, P1, P2, and P3] implies ~T is really a circular argument, for the conclusion is already assumed within the premisses. But, as I said, I was being charitable with the puzzles and not assuming that that was being communicated.
I guess it wasn’t clear, C1 and C2 reffered to the reasonings as well as the conclusions they reached. You say belief is of no importance here, but I don’t see how you can talk about “defeat” if you’re not talking about justified believing.
I’m not sure if I understood what you said here. You agree with what I said in the first bullet or not?
Are you sure that’s correct? If there’s a contradiction within the set of axioms, you could find T and ~T following valid deductions, couldn’t you? Proving ~T and proving that the reasoning leading to T was invalid are only equivalent if you assume the axioms are not contradictory. Am I wrong?
The problem I see here is: it seems like you are assuming that the proof of ~T shows clearly the problem (i.e. the invalid reasoning step) with the proof of T I previously reasoned. If it doesn’t, all the information I have is that both T and ~T are derived apparently validly from the axioms F, P1, P2, and P3. I don’t see why logic would force me to accept ~T instead of believing there’s a mistake I can’t see in the proof Ms. Math showed me, or, more plausibly, to conclude that the axioms are contradictory.
“Defeat” would solely consist in the recognition of admitting to ~T instead of T. Not a matter of belief per se.
No, I don’t.
T cannot be derived from [P1, P2, and P3], but ~T can on account of F serving as a corrective that invalidates T. The only assumptions I’ve made are 1) Ms. Math is not an ivory tower authoritarian and 2) that she wouldn’t be so illogical as to assert a circular argument where F would merely be a premiss, instead of being equivalent to the proper (valid) conclusion ~T.
Anyway, I suppose there’s no more to be said about this, but you can ask for further clarification if you want.
Oh, now I see what you mean. I interpreted F as a new promiss, a new axiom, not a whole argument about the (mistaken) reasoning that proved T. For example, (wikipedia tells me that) the axiom of determinacy is inconsistent with the axiom of choice. If I had proved T in ZFC, and Ms. Math asserted the Axiom of Determinacy and proved ~T in ZFC+AD, and I didn’t know beforehand that AD is inconsistent with AC, I would still need to find out what was the problem.
I still think this is more consistent with the text of the original post, but now I understand what you meant by ” I was being charitable with the puzzles”.
Thank you for you attention.