Isn't proving $A$ iff $B$ iff $C$ by showing $Ato B$ and $Bto C$ and $Cto A$ circular?
$begingroup$
Suppose you have $A$ iff $B$ iff $C$.
If you assume $A$ to be true to prove $B$, $B$ to be true to prove $C$, and $C$ to be true to prove $A$, then doesn't that imply you've assumed $A$ to be true to prove $A$?
I ask because of the method of proof in https://proofwiki.org/wiki/Equivalence_of_Well-Ordering_Principle_and_Induction.
logic
edited Feb 18 at 8:25
Asaf Karagila♦
306k33438769
asked Feb 17 at 20:05
Jossie CalderonJossie Calderon
290111
$endgroup$
add a comment |
$begingroup$
Suppose you have $A$ iff $B$ iff $C$.
If you assume $A$ to be true to prove $B$, $B$ to be true to prove $C$, and $C$ to be true to prove $A$, then doesn't that imply you've assumed $A$ to be true to prove $A$?
I ask because of the method of proof in https://proofwiki.org/wiki/Equivalence_of_Well-Ordering_Principle_and_Induction.
logic
edited Feb 18 at 8:25
Asaf Karagila♦
306k33438769
asked Feb 17 at 20:05
Jossie CalderonJossie Calderon
290111
$endgroup$
13
$begingroup$
You are not proving any of them. You are proving that they are equivalent.
$endgroup$
– Tobias Kildetoft
Feb 17 at 20:10
3
$begingroup$
The real question I have, is why do you need A, B, and C in your question? Isn't just proving that A iff B circular enough for you?
$endgroup$
– Asaf Karagila♦
Feb 18 at 8:25
$begingroup$
@Karagila because of the given context.
$endgroup$
– Jossie Calderon
Feb 19 at 17:39
add a comment |
$begingroup$
Suppose you have $A$ iff $B$ iff $C$.
If you assume $A$ to be true to prove $B$, $B$ to be true to prove $C$, and $C$ to be true to prove $A$, then doesn't that imply you've assumed $A$ to be true to prove $A$?
I ask because of the method of proof in https://proofwiki.org/wiki/Equivalence_of_Well-Ordering_Principle_and_Induction.
logic
edited Feb 18 at 8:25
Asaf Karagila♦
306k33438769
asked Feb 17 at 20:05
Jossie CalderonJossie Calderon
290111
$endgroup$
Suppose you have $A$ iff $B$ iff $C$.
If you assume $A$ to be true to prove $B$, $B$ to be true to prove $C$, and $C$ to be true to prove $A$, then doesn't that imply you've assumed $A$ to be true to prove $A$?
I ask because of the method of proof in https://proofwiki.org/wiki/Equivalence_of_Well-Ordering_Principle_and_Induction.
logic
logic
edited Feb 18 at 8:25
Asaf Karagila♦
306k33438769
asked Feb 17 at 20:05
Jossie CalderonJossie Calderon
290111
edited Feb 18 at 8:25
Asaf Karagila♦
306k33438769
asked Feb 17 at 20:05
Jossie CalderonJossie Calderon
290111
edited Feb 18 at 8:25
Asaf Karagila♦
306k33438769
edited Feb 18 at 8:25
Asaf Karagila♦
306k33438769
edited Feb 18 at 8:25
Asaf Karagila♦
306k33438769
306k33438769
asked Feb 17 at 20:05
Jossie CalderonJossie Calderon
290111
asked Feb 17 at 20:05
Jossie CalderonJossie Calderon
290111
asked Feb 17 at 20:05
Jossie CalderonJossie Calderon
290111
290111
13
$begingroup$
You are not proving any of them. You are proving that they are equivalent.
$endgroup$
– Tobias Kildetoft
Feb 17 at 20:10
3
$begingroup$
The real question I have, is why do you need A, B, and C in your question? Isn't just proving that A iff B circular enough for you?
$endgroup$
– Asaf Karagila♦
Feb 18 at 8:25
$begingroup$
@Karagila because of the given context.
$endgroup$
– Jossie Calderon
Feb 19 at 17:39
add a comment |
13
$begingroup$
You are not proving any of them. You are proving that they are equivalent.
$endgroup$
– Tobias Kildetoft
Feb 17 at 20:10
3
$begingroup$
The real question I have, is why do you need A, B, and C in your question? Isn't just proving that A iff B circular enough for you?
$endgroup$
– Asaf Karagila♦
Feb 18 at 8:25
$begingroup$
@Karagila because of the given context.
$endgroup$
– Jossie Calderon
Feb 19 at 17:39
13
13
$begingroup$
You are not proving any of them. You are proving that they are equivalent.
$endgroup$
– Tobias Kildetoft
Feb 17 at 20:10
$begingroup$
You are not proving any of them. You are proving that they are equivalent.
$endgroup$
– Tobias Kildetoft
Feb 17 at 20:10
3
3
$begingroup$
The real question I have, is why do you need A, B, and C in your question? Isn't just proving that A iff B circular enough for you?
$endgroup$
– Asaf Karagila♦
Feb 18 at 8:25
$begingroup$
The real question I have, is why do you need A, B, and C in your question? Isn't just proving that A iff B circular enough for you?
$endgroup$
– Asaf Karagila♦
Feb 18 at 8:25
$begingroup$
@Karagila because of the given context.
$endgroup$
– Jossie Calderon
Feb 19 at 17:39
$begingroup$
@Karagila because of the given context.
$endgroup$
– Jossie Calderon
Feb 19 at 17:39
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Yes, you are absolutely right the proof $A implies B implies C implies A$ only shows that $A,B,C$ are equivalent. So if one of them is true, all the others are true and if one is false all the others are false. However, the proof you linked doesn't try to show that the principle of mathematical/complete induction is true or the principle of well ordering is true, it only shows that they are equivalent. In fact, in the usual framework of mathematics these are taken as axioms. The proof shows that you only need to assume one of them as an axiom and you get all the others for free.
answered Feb 17 at 20:11
Jannik PittJannik Pitt
646517
$endgroup$
$begingroup$
Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
$endgroup$
– Remellion
Feb 18 at 2:49
add a comment |
$begingroup$
Yes. But, this helps math, it gives a bunch of statements, that we either can all dismiss at once with a disproof of one, or accept if we accept or prove the others. Goldbach's conjecture, is sufficient to proves Bertrand's postulate. But, Bertrand's postulate is not sufficient to prove Goldbach's conjecture. In this case, we can't use the proof of Bertrand's postulate to prove Goldbach's conjecture. But we could use ANY proof of Goldbach's conjecture, to prove Bertrand's postulate again possibly in a new way. IFF means we could prove any statement in the chain and the other's follow.
answered Feb 23 at 19:34
Roddy MacPheeRoddy MacPhee
473117
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Yes, you are absolutely right the proof $A implies B implies C implies A$ only shows that $A,B,C$ are equivalent. So if one of them is true, all the others are true and if one is false all the others are false. However, the proof you linked doesn't try to show that the principle of mathematical/complete induction is true or the principle of well ordering is true, it only shows that they are equivalent. In fact, in the usual framework of mathematics these are taken as axioms. The proof shows that you only need to assume one of them as an axiom and you get all the others for free.
answered Feb 17 at 20:11
Jannik PittJannik Pitt
646517
$endgroup$
$begingroup$
Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
$endgroup$
– Remellion
Feb 18 at 2:49
add a comment |
$begingroup$
Yes, you are absolutely right the proof $A implies B implies C implies A$ only shows that $A,B,C$ are equivalent. So if one of them is true, all the others are true and if one is false all the others are false. However, the proof you linked doesn't try to show that the principle of mathematical/complete induction is true or the principle of well ordering is true, it only shows that they are equivalent. In fact, in the usual framework of mathematics these are taken as axioms. The proof shows that you only need to assume one of them as an axiom and you get all the others for free.
answered Feb 17 at 20:11
Jannik PittJannik Pitt
646517
$endgroup$
$begingroup$
Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
$endgroup$
– Remellion
Feb 18 at 2:49
add a comment |
$begingroup$
Yes, you are absolutely right the proof $A implies B implies C implies A$ only shows that $A,B,C$ are equivalent. So if one of them is true, all the others are true and if one is false all the others are false. However, the proof you linked doesn't try to show that the principle of mathematical/complete induction is true or the principle of well ordering is true, it only shows that they are equivalent. In fact, in the usual framework of mathematics these are taken as axioms. The proof shows that you only need to assume one of them as an axiom and you get all the others for free.
answered Feb 17 at 20:11
Jannik PittJannik Pitt
646517
$endgroup$
Yes, you are absolutely right the proof $A implies B implies C implies A$ only shows that $A,B,C$ are equivalent. So if one of them is true, all the others are true and if one is false all the others are false. However, the proof you linked doesn't try to show that the principle of mathematical/complete induction is true or the principle of well ordering is true, it only shows that they are equivalent. In fact, in the usual framework of mathematics these are taken as axioms. The proof shows that you only need to assume one of them as an axiom and you get all the others for free.
answered Feb 17 at 20:11
Jannik PittJannik Pitt
646517
answered Feb 17 at 20:11
Jannik PittJannik Pitt
646517
answered Feb 17 at 20:11
Jannik PittJannik Pitt
646517
answered Feb 17 at 20:11
Jannik PittJannik Pitt
646517
646517
$begingroup$
Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
$endgroup$
– Remellion
Feb 18 at 2:49
add a comment |
$begingroup$
Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
$endgroup$
– Remellion
Feb 18 at 2:49
$begingroup$
Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
$endgroup$
– Remellion
Feb 18 at 2:49
$begingroup$
Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
$endgroup$
– Remellion
Feb 18 at 2:49
add a comment |
$begingroup$
Yes. But, this helps math, it gives a bunch of statements, that we either can all dismiss at once with a disproof of one, or accept if we accept or prove the others. Goldbach's conjecture, is sufficient to proves Bertrand's postulate. But, Bertrand's postulate is not sufficient to prove Goldbach's conjecture. In this case, we can't use the proof of Bertrand's postulate to prove Goldbach's conjecture. But we could use ANY proof of Goldbach's conjecture, to prove Bertrand's postulate again possibly in a new way. IFF means we could prove any statement in the chain and the other's follow.
answered Feb 23 at 19:34
Roddy MacPheeRoddy MacPhee
473117
$endgroup$
add a comment |
$begingroup$
Yes. But, this helps math, it gives a bunch of statements, that we either can all dismiss at once with a disproof of one, or accept if we accept or prove the others. Goldbach's conjecture, is sufficient to proves Bertrand's postulate. But, Bertrand's postulate is not sufficient to prove Goldbach's conjecture. In this case, we can't use the proof of Bertrand's postulate to prove Goldbach's conjecture. But we could use ANY proof of Goldbach's conjecture, to prove Bertrand's postulate again possibly in a new way. IFF means we could prove any statement in the chain and the other's follow.
answered Feb 23 at 19:34
Roddy MacPheeRoddy MacPhee
473117
$endgroup$
add a comment |
$begingroup$
Yes. But, this helps math, it gives a bunch of statements, that we either can all dismiss at once with a disproof of one, or accept if we accept or prove the others. Goldbach's conjecture, is sufficient to proves Bertrand's postulate. But, Bertrand's postulate is not sufficient to prove Goldbach's conjecture. In this case, we can't use the proof of Bertrand's postulate to prove Goldbach's conjecture. But we could use ANY proof of Goldbach's conjecture, to prove Bertrand's postulate again possibly in a new way. IFF means we could prove any statement in the chain and the other's follow.
answered Feb 23 at 19:34
Roddy MacPheeRoddy MacPhee
473117
$endgroup$
Yes. But, this helps math, it gives a bunch of statements, that we either can all dismiss at once with a disproof of one, or accept if we accept or prove the others. Goldbach's conjecture, is sufficient to proves Bertrand's postulate. But, Bertrand's postulate is not sufficient to prove Goldbach's conjecture. In this case, we can't use the proof of Bertrand's postulate to prove Goldbach's conjecture. But we could use ANY proof of Goldbach's conjecture, to prove Bertrand's postulate again possibly in a new way. IFF means we could prove any statement in the chain and the other's follow.
answered Feb 23 at 19:34
Roddy MacPheeRoddy MacPhee
473117
answered Feb 23 at 19:34
Roddy MacPheeRoddy MacPhee
473117
answered Feb 23 at 19:34
Roddy MacPheeRoddy MacPhee
473117
answered Feb 23 at 19:34
Roddy MacPheeRoddy MacPhee
473117
473117
add a comment |
add a comment |
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13
$begingroup$
You are not proving any of them. You are proving that they are equivalent.
$endgroup$
– Tobias Kildetoft
Feb 17 at 20:10
3
$begingroup$
The real question I have, is why do you need A, B, and C in your question? Isn't just proving that A iff B circular enough for you?
$endgroup$
– Asaf Karagila♦
Feb 18 at 8:25
$begingroup$
@Karagila because of the given context.
$endgroup$
– Jossie Calderon
Feb 19 at 17:39