Isn't this circular logic?
$begingroup$
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
asked 8 hours ago
Jossie CalderonJossie Calderon
18911
$endgroup$
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
asked 8 hours ago
Jossie CalderonJossie Calderon
18911
$endgroup$
7
$begingroup$
You are not proving any of them. You are proving that they are equivalent.
$endgroup$
– Tobias Kildetoft
7 hours ago
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
asked 8 hours ago
Jossie CalderonJossie Calderon
18911
$endgroup$
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
asked 8 hours ago
Jossie CalderonJossie Calderon
18911
asked 8 hours ago
Jossie CalderonJossie Calderon
18911
asked 8 hours ago
Jossie CalderonJossie Calderon
18911
asked 8 hours ago
Jossie CalderonJossie Calderon
18911
asked 8 hours ago
Jossie CalderonJossie Calderon
18911
18911
7
$begingroup$
You are not proving any of them. You are proving that they are equivalent.
$endgroup$
– Tobias Kildetoft
7 hours ago
add a comment |
7
$begingroup$
You are not proving any of them. You are proving that they are equivalent.
$endgroup$
– Tobias Kildetoft
7 hours ago
7
7
$begingroup$
You are not proving any of them. You are proving that they are equivalent.
$endgroup$
– Tobias Kildetoft
7 hours ago
$begingroup$
You are not proving any of them. You are proving that they are equivalent.
$endgroup$
– Tobias Kildetoft
7 hours ago
add a comment |
1 Answer
1
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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 7 hours ago
Jannik PittJannik Pitt
573517
$endgroup$
$begingroup$
Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
$endgroup$
– Remellion
1 hour ago
add a comment |
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1 Answer
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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 7 hours ago
Jannik PittJannik Pitt
573517
$endgroup$
$begingroup$
Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
$endgroup$
– Remellion
1 hour ago
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 7 hours ago
Jannik PittJannik Pitt
573517
$endgroup$
$begingroup$
Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
$endgroup$
– Remellion
1 hour ago
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 7 hours ago
Jannik PittJannik Pitt
573517
$endgroup$
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 7 hours ago
Jannik PittJannik Pitt
573517
answered 7 hours ago
Jannik PittJannik Pitt
573517
answered 7 hours ago
Jannik PittJannik Pitt
573517
answered 7 hours ago
Jannik PittJannik Pitt
573517
573517
$begingroup$
Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
$endgroup$
– Remellion
1 hour ago
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
1 hour ago
$begingroup$
Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
$endgroup$
– Remellion
1 hour ago
$begingroup$
Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
$endgroup$
– Remellion
1 hour ago
add a comment |
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7
$begingroup$
You are not proving any of them. You are proving that they are equivalent.
$endgroup$
– Tobias Kildetoft
7 hours ago