Logic - Member of an inductive set that is not a member of the set of theorems in a formal system?

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First question on this site. Hope to ask/answer many more in the future.



I'm currently self-studying An Introduction to Mathematical Logic by Richard E. Hodel and came across an interesting exercise. This is how it is introduced:




Let F be a formal system, let FOR be the set of formulas of F, and let THM be the set of theorems of F. A subset I of FOR is said to be inductive if it satisfies these two conditions: (a) Every axiom of the formal system F is in I; (b) If A1, ... , An/B is a rule of inference of F, and each of the hypotheses A1, ... , An are in I, then B is also in I.




The exercise leads one to prove that THM is a subset of any arbitrary inductive set I, and that THM is the smallest inductive subset of FOR, which I already worked out successfully.



The exercise does not imply, however, that THM = I.



So my question is then, what would be an example of a member of an inductive set that is not a member of THM? And, more generally, when does a formula fail to be a theorem, but succeed in being a member of I?



I'm still on chapter 1 of the book, so maybe my question will be answered down the line, but I am still curious and think that this would help me understand the definitions of a formal system. So any help is appreciated.



Thanks




elementary-set-theory logic induction formal-systems




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  • 3




    All of FOR forms an inductive set; but if F is consistent, then it is not equal to THM.
    – Daniel Schepler
    Aug 14 at 1:51














up vote
3
down vote

favorite












First question on this site. Hope to ask/answer many more in the future.



I'm currently self-studying An Introduction to Mathematical Logic by Richard E. Hodel and came across an interesting exercise. This is how it is introduced:




Let F be a formal system, let FOR be the set of formulas of F, and let THM be the set of theorems of F. A subset I of FOR is said to be inductive if it satisfies these two conditions: (a) Every axiom of the formal system F is in I; (b) If A1, ... , An/B is a rule of inference of F, and each of the hypotheses A1, ... , An are in I, then B is also in I.




The exercise leads one to prove that THM is a subset of any arbitrary inductive set I, and that THM is the smallest inductive subset of FOR, which I already worked out successfully.



The exercise does not imply, however, that THM = I.



So my question is then, what would be an example of a member of an inductive set that is not a member of THM? And, more generally, when does a formula fail to be a theorem, but succeed in being a member of I?



I'm still on chapter 1 of the book, so maybe my question will be answered down the line, but I am still curious and think that this would help me understand the definitions of a formal system. So any help is appreciated.



Thanks




elementary-set-theory logic induction formal-systems




share|cite|improve this question
















  • 3




    All of FOR forms an inductive set; but if F is consistent, then it is not equal to THM.
    – Daniel Schepler
    Aug 14 at 1:51












up vote
3
down vote

favorite









up vote
3
down vote

favorite











First question on this site. Hope to ask/answer many more in the future.



I'm currently self-studying An Introduction to Mathematical Logic by Richard E. Hodel and came across an interesting exercise. This is how it is introduced:




Let F be a formal system, let FOR be the set of formulas of F, and let THM be the set of theorems of F. A subset I of FOR is said to be inductive if it satisfies these two conditions: (a) Every axiom of the formal system F is in I; (b) If A1, ... , An/B is a rule of inference of F, and each of the hypotheses A1, ... , An are in I, then B is also in I.




The exercise leads one to prove that THM is a subset of any arbitrary inductive set I, and that THM is the smallest inductive subset of FOR, which I already worked out successfully.



The exercise does not imply, however, that THM = I.



So my question is then, what would be an example of a member of an inductive set that is not a member of THM? And, more generally, when does a formula fail to be a theorem, but succeed in being a member of I?



I'm still on chapter 1 of the book, so maybe my question will be answered down the line, but I am still curious and think that this would help me understand the definitions of a formal system. So any help is appreciated.



Thanks




elementary-set-theory logic induction formal-systems




share|cite|improve this question












First question on this site. Hope to ask/answer many more in the future.



I'm currently self-studying An Introduction to Mathematical Logic by Richard E. Hodel and came across an interesting exercise. This is how it is introduced:




Let F be a formal system, let FOR be the set of formulas of F, and let THM be the set of theorems of F. A subset I of FOR is said to be inductive if it satisfies these two conditions: (a) Every axiom of the formal system F is in I; (b) If A1, ... , An/B is a rule of inference of F, and each of the hypotheses A1, ... , An are in I, then B is also in I.




The exercise leads one to prove that THM is a subset of any arbitrary inductive set I, and that THM is the smallest inductive subset of FOR, which I already worked out successfully.



The exercise does not imply, however, that THM = I.



So my question is then, what would be an example of a member of an inductive set that is not a member of THM? And, more generally, when does a formula fail to be a theorem, but succeed in being a member of I?



I'm still on chapter 1 of the book, so maybe my question will be answered down the line, but I am still curious and think that this would help me understand the definitions of a formal system. So any help is appreciated.



Thanks





elementary-set-theory logic induction formal-systems





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share|cite|improve this question




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asked Aug 14 at 1:46









mraaron

183




183







  • 3




    All of FOR forms an inductive set; but if F is consistent, then it is not equal to THM.
    – Daniel Schepler
    Aug 14 at 1:51












  • 3




    All of FOR forms an inductive set; but if F is consistent, then it is not equal to THM.
    – Daniel Schepler
    Aug 14 at 1:51







3




3




All of FOR forms an inductive set; but if F is consistent, then it is not equal to THM.
– Daniel Schepler
Aug 14 at 1:51




All of FOR forms an inductive set; but if F is consistent, then it is not equal to THM.
– Daniel Schepler
Aug 14 at 1:51










2 Answers
2






active

oldest

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up vote
5
down vote



accepted










An example of an inductive set is the set FOR of all formulas. This is strictly larger than THM in most cases of interest. (I see I've just repeated Daniel Schepler's comment here.)



More generally, suppose that not every formula is a theorem, i.e. $textTHMsubsetneqtextFOR$. Then pick any formula $varphiin textFORsetminus textTHM$, and add it to $F$ as an additional axiom, obtaining a new formal system $F'$. Then the set $textTHM'$ of theorems of $F'$ is an inductive set which is larger than $textTHM$, since it includes $varphi$.



In fact, you can add any set of formulas to $F$ as new axioms, and the resulting set of theorems will be an inductive set. The case of $textFOR$ occurs when you take every formula as an axiom, or just enough axioms to make the system inconsistent.






share|cite|improve this answer




















  • Thanks for the insight. This certainly helped clear things up for me
    – mraaron
    Aug 15 at 11:22

















up vote
3
down vote













Suppose we have any fixed model $M$ of $F$. Then the set of formulas $phi$ such that $M models phi$ is an inductive set. However, unless $F$ is complete then this set will include formulas not in THM.






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  • Thanks for the great and succinct answer
    – mraaron
    Aug 15 at 11:24










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote



accepted










An example of an inductive set is the set FOR of all formulas. This is strictly larger than THM in most cases of interest. (I see I've just repeated Daniel Schepler's comment here.)



More generally, suppose that not every formula is a theorem, i.e. $textTHMsubsetneqtextFOR$. Then pick any formula $varphiin textFORsetminus textTHM$, and add it to $F$ as an additional axiom, obtaining a new formal system $F'$. Then the set $textTHM'$ of theorems of $F'$ is an inductive set which is larger than $textTHM$, since it includes $varphi$.



In fact, you can add any set of formulas to $F$ as new axioms, and the resulting set of theorems will be an inductive set. The case of $textFOR$ occurs when you take every formula as an axiom, or just enough axioms to make the system inconsistent.






share|cite|improve this answer




















  • Thanks for the insight. This certainly helped clear things up for me
    – mraaron
    Aug 15 at 11:22














up vote
5
down vote



accepted










An example of an inductive set is the set FOR of all formulas. This is strictly larger than THM in most cases of interest. (I see I've just repeated Daniel Schepler's comment here.)



More generally, suppose that not every formula is a theorem, i.e. $textTHMsubsetneqtextFOR$. Then pick any formula $varphiin textFORsetminus textTHM$, and add it to $F$ as an additional axiom, obtaining a new formal system $F'$. Then the set $textTHM'$ of theorems of $F'$ is an inductive set which is larger than $textTHM$, since it includes $varphi$.



In fact, you can add any set of formulas to $F$ as new axioms, and the resulting set of theorems will be an inductive set. The case of $textFOR$ occurs when you take every formula as an axiom, or just enough axioms to make the system inconsistent.






share|cite|improve this answer




















  • Thanks for the insight. This certainly helped clear things up for me
    – mraaron
    Aug 15 at 11:22












up vote
5
down vote



accepted







up vote
5
down vote



accepted






An example of an inductive set is the set FOR of all formulas. This is strictly larger than THM in most cases of interest. (I see I've just repeated Daniel Schepler's comment here.)



More generally, suppose that not every formula is a theorem, i.e. $textTHMsubsetneqtextFOR$. Then pick any formula $varphiin textFORsetminus textTHM$, and add it to $F$ as an additional axiom, obtaining a new formal system $F'$. Then the set $textTHM'$ of theorems of $F'$ is an inductive set which is larger than $textTHM$, since it includes $varphi$.



In fact, you can add any set of formulas to $F$ as new axioms, and the resulting set of theorems will be an inductive set. The case of $textFOR$ occurs when you take every formula as an axiom, or just enough axioms to make the system inconsistent.






share|cite|improve this answer












An example of an inductive set is the set FOR of all formulas. This is strictly larger than THM in most cases of interest. (I see I've just repeated Daniel Schepler's comment here.)



More generally, suppose that not every formula is a theorem, i.e. $textTHMsubsetneqtextFOR$. Then pick any formula $varphiin textFORsetminus textTHM$, and add it to $F$ as an additional axiom, obtaining a new formal system $F'$. Then the set $textTHM'$ of theorems of $F'$ is an inductive set which is larger than $textTHM$, since it includes $varphi$.



In fact, you can add any set of formulas to $F$ as new axioms, and the resulting set of theorems will be an inductive set. The case of $textFOR$ occurs when you take every formula as an axiom, or just enough axioms to make the system inconsistent.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 14 at 1:57









Alex Kruckman

23.6k22453




23.6k22453











  • Thanks for the insight. This certainly helped clear things up for me
    – mraaron
    Aug 15 at 11:22
















  • Thanks for the insight. This certainly helped clear things up for me
    – mraaron
    Aug 15 at 11:22















Thanks for the insight. This certainly helped clear things up for me
– mraaron
Aug 15 at 11:22




Thanks for the insight. This certainly helped clear things up for me
– mraaron
Aug 15 at 11:22










up vote
3
down vote













Suppose we have any fixed model $M$ of $F$. Then the set of formulas $phi$ such that $M models phi$ is an inductive set. However, unless $F$ is complete then this set will include formulas not in THM.






share|cite|improve this answer




















  • Thanks for the great and succinct answer
    – mraaron
    Aug 15 at 11:24














up vote
3
down vote













Suppose we have any fixed model $M$ of $F$. Then the set of formulas $phi$ such that $M models phi$ is an inductive set. However, unless $F$ is complete then this set will include formulas not in THM.






share|cite|improve this answer




















  • Thanks for the great and succinct answer
    – mraaron
    Aug 15 at 11:24












up vote
3
down vote










up vote
3
down vote









Suppose we have any fixed model $M$ of $F$. Then the set of formulas $phi$ such that $M models phi$ is an inductive set. However, unless $F$ is complete then this set will include formulas not in THM.






share|cite|improve this answer












Suppose we have any fixed model $M$ of $F$. Then the set of formulas $phi$ such that $M models phi$ is an inductive set. However, unless $F$ is complete then this set will include formulas not in THM.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 14 at 2:33









Daniel Schepler

7,0381514




7,0381514











  • Thanks for the great and succinct answer
    – mraaron
    Aug 15 at 11:24
















  • Thanks for the great and succinct answer
    – mraaron
    Aug 15 at 11:24















Thanks for the great and succinct answer
– mraaron
Aug 15 at 11:24




Thanks for the great and succinct answer
– mraaron
Aug 15 at 11:24

















 

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