Isn't this circular logic?












5












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 10




    $begingroup$
    You are not proving any of them. You are proving that they are equivalent.
    $endgroup$
    – Tobias Kildetoft
    9 hours ago
















5












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 10




    $begingroup$
    You are not proving any of them. You are proving that they are equivalent.
    $endgroup$
    – Tobias Kildetoft
    9 hours ago














5












5








5


1



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 9 hours ago









Jossie CalderonJossie Calderon

19911




19911








  • 10




    $begingroup$
    You are not proving any of them. You are proving that they are equivalent.
    $endgroup$
    – Tobias Kildetoft
    9 hours ago














  • 10




    $begingroup$
    You are not proving any of them. You are proving that they are equivalent.
    $endgroup$
    – Tobias Kildetoft
    9 hours ago








10




10




$begingroup$
You are not proving any of them. You are proving that they are equivalent.
$endgroup$
– Tobias Kildetoft
9 hours ago




$begingroup$
You are not proving any of them. You are proving that they are equivalent.
$endgroup$
– Tobias Kildetoft
9 hours ago










1 Answer
1






active

oldest

votes


















16












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
    $endgroup$
    – Remellion
    2 hours ago













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1 Answer
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oldest

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1 Answer
1






active

oldest

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active

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active

oldest

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16












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
    $endgroup$
    – Remellion
    2 hours ago


















16












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
    $endgroup$
    – Remellion
    2 hours ago
















16












16








16





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 9 hours ago









Jannik PittJannik Pitt

593517




593517












  • $begingroup$
    Or that if you prove from other things one of $A$, $B$ or $C$ (usually the easiest one), the others follow immediately.
    $endgroup$
    – Remellion
    2 hours 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
    2 hours 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
2 hours 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
2 hours ago




















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