Independence and Consistence of Formal Systems

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Let $S$ be a formal system with axioms $A,A_1,dots,A_n$. The system $S$ is said to be consistent if no contradiction can be proved (i.e. we can’t prove both a formula and its negation). If $S$ is consistent, the axiom $A$ is said to be dependent from $A_1,dots,A_n$ if $A$ can be proved using only $A_1,dots,A_n$.



Given these definitions, if $S$ and $S'$ are both consistent, with $S’$ created from $S$ by negating $A$ (i.e. the axioms of $S'$ are $lnot A,A_1, dots,A_n$), then obviously $A$ is not dependent from $A_1,dots,A_n$.



Does the opposite holds? That is can we prove that if $S$ is consistent ad $S’$ is not consistent then $A$ can be derived from $A_1,dots,A_n$?




logic first-order-logic proof-theory formal-systems




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    Let $S$ be a formal system with axioms $A,A_1,dots,A_n$. The system $S$ is said to be consistent if no contradiction can be proved (i.e. we can’t prove both a formula and its negation). If $S$ is consistent, the axiom $A$ is said to be dependent from $A_1,dots,A_n$ if $A$ can be proved using only $A_1,dots,A_n$.



    Given these definitions, if $S$ and $S'$ are both consistent, with $S’$ created from $S$ by negating $A$ (i.e. the axioms of $S'$ are $lnot A,A_1, dots,A_n$), then obviously $A$ is not dependent from $A_1,dots,A_n$.



    Does the opposite holds? That is can we prove that if $S$ is consistent ad $S’$ is not consistent then $A$ can be derived from $A_1,dots,A_n$?




    logic first-order-logic proof-theory formal-systems




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      Let $S$ be a formal system with axioms $A,A_1,dots,A_n$. The system $S$ is said to be consistent if no contradiction can be proved (i.e. we can’t prove both a formula and its negation). If $S$ is consistent, the axiom $A$ is said to be dependent from $A_1,dots,A_n$ if $A$ can be proved using only $A_1,dots,A_n$.



      Given these definitions, if $S$ and $S'$ are both consistent, with $S’$ created from $S$ by negating $A$ (i.e. the axioms of $S'$ are $lnot A,A_1, dots,A_n$), then obviously $A$ is not dependent from $A_1,dots,A_n$.



      Does the opposite holds? That is can we prove that if $S$ is consistent ad $S’$ is not consistent then $A$ can be derived from $A_1,dots,A_n$?




      logic first-order-logic proof-theory formal-systems




      share|cite|improve this question













      Let $S$ be a formal system with axioms $A,A_1,dots,A_n$. The system $S$ is said to be consistent if no contradiction can be proved (i.e. we can’t prove both a formula and its negation). If $S$ is consistent, the axiom $A$ is said to be dependent from $A_1,dots,A_n$ if $A$ can be proved using only $A_1,dots,A_n$.



      Given these definitions, if $S$ and $S'$ are both consistent, with $S’$ created from $S$ by negating $A$ (i.e. the axioms of $S'$ are $lnot A,A_1, dots,A_n$), then obviously $A$ is not dependent from $A_1,dots,A_n$.



      Does the opposite holds? That is can we prove that if $S$ is consistent ad $S’$ is not consistent then $A$ can be derived from $A_1,dots,A_n$?





      logic first-order-logic proof-theory formal-systems





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      share|cite|improve this question




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      edited Jul 28 at 8:03









      Taroccoesbrocco

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      asked Jul 28 at 7:21









      Markus Steiner

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          The answer is positive, at least in first-order classical logic (which is the logic you refer to, I guess), where you have completeness and soundness theorems.



          Completeness theorem states that if every model of $A_1, dots, A_n$ is a model of $A$ then $A$ can be derived from $A_1, dots, A_n$.



          Soundness theorem is the converse of completeness, it states that if $A$ can be derived from $A_1, dots, A_n$ then every model of $A_1, dots, A_n$ is a model of $A$.



          If $S' = A_1, dots, A_n, lnot A$ is not consistent then $S'$ is not satisfiable i.e. there is no model satisfying $A_1, dots, A_n, lnot A$ (because by soundness theorem, if there were such a model, it would satisfy a contradiction, that is impossible). So, every model of $A_1, dots, A_n$ cannot be a model of $lnot A$ and hence it is a model of $A$. Therefore, $A$ is derivable from $A_1, dots, A_n$ by completeness theorem.



          Note that the hypothesis that $S$ is consistent is superfluous.






          share|cite|improve this answer























          • So we need to add an essential completeness hypothesis on S?
            – Markus Steiner
            Jul 28 at 8:18










          • @MarkusSteiner - When you talk about a formal system $S$, what do you refer to? I guess it is a set of axioms (in a given language) that you add to a (complete and sound) derivation system for first-order classical logic, such as Hilbert system, natural deduction or sequent calculus. Isn't it?
            – Taroccoesbrocco
            Jul 28 at 8:42











          • @MarkusSteiner Your comment suggests to me that you might be confusing completeness of an axiom system (which means that, for any sentence $A$, either $A$ or $neg A$ is deducible from those axioms) with the completeness theorem for first-order logic (which says that every semantic consequence of any axiom system is deducible from that axiom system). It is unfortunate that these two different concepts have very similar names, both involving the word "completeness".
            – Andreas Blass
            Jul 28 at 13:05










          • Yes you are right, I misunderstood, anyway now it's clear, thank you for you answer.
            – Markus Steiner
            Jul 28 at 17:21










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          The answer is positive, at least in first-order classical logic (which is the logic you refer to, I guess), where you have completeness and soundness theorems.



          Completeness theorem states that if every model of $A_1, dots, A_n$ is a model of $A$ then $A$ can be derived from $A_1, dots, A_n$.



          Soundness theorem is the converse of completeness, it states that if $A$ can be derived from $A_1, dots, A_n$ then every model of $A_1, dots, A_n$ is a model of $A$.



          If $S' = A_1, dots, A_n, lnot A$ is not consistent then $S'$ is not satisfiable i.e. there is no model satisfying $A_1, dots, A_n, lnot A$ (because by soundness theorem, if there were such a model, it would satisfy a contradiction, that is impossible). So, every model of $A_1, dots, A_n$ cannot be a model of $lnot A$ and hence it is a model of $A$. Therefore, $A$ is derivable from $A_1, dots, A_n$ by completeness theorem.



          Note that the hypothesis that $S$ is consistent is superfluous.






          share|cite|improve this answer























          • So we need to add an essential completeness hypothesis on S?
            – Markus Steiner
            Jul 28 at 8:18










          • @MarkusSteiner - When you talk about a formal system $S$, what do you refer to? I guess it is a set of axioms (in a given language) that you add to a (complete and sound) derivation system for first-order classical logic, such as Hilbert system, natural deduction or sequent calculus. Isn't it?
            – Taroccoesbrocco
            Jul 28 at 8:42











          • @MarkusSteiner Your comment suggests to me that you might be confusing completeness of an axiom system (which means that, for any sentence $A$, either $A$ or $neg A$ is deducible from those axioms) with the completeness theorem for first-order logic (which says that every semantic consequence of any axiom system is deducible from that axiom system). It is unfortunate that these two different concepts have very similar names, both involving the word "completeness".
            – Andreas Blass
            Jul 28 at 13:05










          • Yes you are right, I misunderstood, anyway now it's clear, thank you for you answer.
            – Markus Steiner
            Jul 28 at 17:21














          up vote
          0
          down vote



          accepted










          The answer is positive, at least in first-order classical logic (which is the logic you refer to, I guess), where you have completeness and soundness theorems.



          Completeness theorem states that if every model of $A_1, dots, A_n$ is a model of $A$ then $A$ can be derived from $A_1, dots, A_n$.



          Soundness theorem is the converse of completeness, it states that if $A$ can be derived from $A_1, dots, A_n$ then every model of $A_1, dots, A_n$ is a model of $A$.



          If $S' = A_1, dots, A_n, lnot A$ is not consistent then $S'$ is not satisfiable i.e. there is no model satisfying $A_1, dots, A_n, lnot A$ (because by soundness theorem, if there were such a model, it would satisfy a contradiction, that is impossible). So, every model of $A_1, dots, A_n$ cannot be a model of $lnot A$ and hence it is a model of $A$. Therefore, $A$ is derivable from $A_1, dots, A_n$ by completeness theorem.



          Note that the hypothesis that $S$ is consistent is superfluous.






          share|cite|improve this answer























          • So we need to add an essential completeness hypothesis on S?
            – Markus Steiner
            Jul 28 at 8:18










          • @MarkusSteiner - When you talk about a formal system $S$, what do you refer to? I guess it is a set of axioms (in a given language) that you add to a (complete and sound) derivation system for first-order classical logic, such as Hilbert system, natural deduction or sequent calculus. Isn't it?
            – Taroccoesbrocco
            Jul 28 at 8:42











          • @MarkusSteiner Your comment suggests to me that you might be confusing completeness of an axiom system (which means that, for any sentence $A$, either $A$ or $neg A$ is deducible from those axioms) with the completeness theorem for first-order logic (which says that every semantic consequence of any axiom system is deducible from that axiom system). It is unfortunate that these two different concepts have very similar names, both involving the word "completeness".
            – Andreas Blass
            Jul 28 at 13:05










          • Yes you are right, I misunderstood, anyway now it's clear, thank you for you answer.
            – Markus Steiner
            Jul 28 at 17:21












          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          The answer is positive, at least in first-order classical logic (which is the logic you refer to, I guess), where you have completeness and soundness theorems.



          Completeness theorem states that if every model of $A_1, dots, A_n$ is a model of $A$ then $A$ can be derived from $A_1, dots, A_n$.



          Soundness theorem is the converse of completeness, it states that if $A$ can be derived from $A_1, dots, A_n$ then every model of $A_1, dots, A_n$ is a model of $A$.



          If $S' = A_1, dots, A_n, lnot A$ is not consistent then $S'$ is not satisfiable i.e. there is no model satisfying $A_1, dots, A_n, lnot A$ (because by soundness theorem, if there were such a model, it would satisfy a contradiction, that is impossible). So, every model of $A_1, dots, A_n$ cannot be a model of $lnot A$ and hence it is a model of $A$. Therefore, $A$ is derivable from $A_1, dots, A_n$ by completeness theorem.



          Note that the hypothesis that $S$ is consistent is superfluous.






          share|cite|improve this answer















          The answer is positive, at least in first-order classical logic (which is the logic you refer to, I guess), where you have completeness and soundness theorems.



          Completeness theorem states that if every model of $A_1, dots, A_n$ is a model of $A$ then $A$ can be derived from $A_1, dots, A_n$.



          Soundness theorem is the converse of completeness, it states that if $A$ can be derived from $A_1, dots, A_n$ then every model of $A_1, dots, A_n$ is a model of $A$.



          If $S' = A_1, dots, A_n, lnot A$ is not consistent then $S'$ is not satisfiable i.e. there is no model satisfying $A_1, dots, A_n, lnot A$ (because by soundness theorem, if there were such a model, it would satisfy a contradiction, that is impossible). So, every model of $A_1, dots, A_n$ cannot be a model of $lnot A$ and hence it is a model of $A$. Therefore, $A$ is derivable from $A_1, dots, A_n$ by completeness theorem.



          Note that the hypothesis that $S$ is consistent is superfluous.







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Jul 28 at 8:38


























          answered Jul 28 at 8:13









          Taroccoesbrocco

          3,36941331




          3,36941331











          • So we need to add an essential completeness hypothesis on S?
            – Markus Steiner
            Jul 28 at 8:18










          • @MarkusSteiner - When you talk about a formal system $S$, what do you refer to? I guess it is a set of axioms (in a given language) that you add to a (complete and sound) derivation system for first-order classical logic, such as Hilbert system, natural deduction or sequent calculus. Isn't it?
            – Taroccoesbrocco
            Jul 28 at 8:42











          • @MarkusSteiner Your comment suggests to me that you might be confusing completeness of an axiom system (which means that, for any sentence $A$, either $A$ or $neg A$ is deducible from those axioms) with the completeness theorem for first-order logic (which says that every semantic consequence of any axiom system is deducible from that axiom system). It is unfortunate that these two different concepts have very similar names, both involving the word "completeness".
            – Andreas Blass
            Jul 28 at 13:05










          • Yes you are right, I misunderstood, anyway now it's clear, thank you for you answer.
            – Markus Steiner
            Jul 28 at 17:21
















          • So we need to add an essential completeness hypothesis on S?
            – Markus Steiner
            Jul 28 at 8:18










          • @MarkusSteiner - When you talk about a formal system $S$, what do you refer to? I guess it is a set of axioms (in a given language) that you add to a (complete and sound) derivation system for first-order classical logic, such as Hilbert system, natural deduction or sequent calculus. Isn't it?
            – Taroccoesbrocco
            Jul 28 at 8:42











          • @MarkusSteiner Your comment suggests to me that you might be confusing completeness of an axiom system (which means that, for any sentence $A$, either $A$ or $neg A$ is deducible from those axioms) with the completeness theorem for first-order logic (which says that every semantic consequence of any axiom system is deducible from that axiom system). It is unfortunate that these two different concepts have very similar names, both involving the word "completeness".
            – Andreas Blass
            Jul 28 at 13:05










          • Yes you are right, I misunderstood, anyway now it's clear, thank you for you answer.
            – Markus Steiner
            Jul 28 at 17:21















          So we need to add an essential completeness hypothesis on S?
          – Markus Steiner
          Jul 28 at 8:18




          So we need to add an essential completeness hypothesis on S?
          – Markus Steiner
          Jul 28 at 8:18












          @MarkusSteiner - When you talk about a formal system $S$, what do you refer to? I guess it is a set of axioms (in a given language) that you add to a (complete and sound) derivation system for first-order classical logic, such as Hilbert system, natural deduction or sequent calculus. Isn't it?
          – Taroccoesbrocco
          Jul 28 at 8:42





          @MarkusSteiner - When you talk about a formal system $S$, what do you refer to? I guess it is a set of axioms (in a given language) that you add to a (complete and sound) derivation system for first-order classical logic, such as Hilbert system, natural deduction or sequent calculus. Isn't it?
          – Taroccoesbrocco
          Jul 28 at 8:42













          @MarkusSteiner Your comment suggests to me that you might be confusing completeness of an axiom system (which means that, for any sentence $A$, either $A$ or $neg A$ is deducible from those axioms) with the completeness theorem for first-order logic (which says that every semantic consequence of any axiom system is deducible from that axiom system). It is unfortunate that these two different concepts have very similar names, both involving the word "completeness".
          – Andreas Blass
          Jul 28 at 13:05




          @MarkusSteiner Your comment suggests to me that you might be confusing completeness of an axiom system (which means that, for any sentence $A$, either $A$ or $neg A$ is deducible from those axioms) with the completeness theorem for first-order logic (which says that every semantic consequence of any axiom system is deducible from that axiom system). It is unfortunate that these two different concepts have very similar names, both involving the word "completeness".
          – Andreas Blass
          Jul 28 at 13:05












          Yes you are right, I misunderstood, anyway now it's clear, thank you for you answer.
          – Markus Steiner
          Jul 28 at 17:21




          Yes you are right, I misunderstood, anyway now it's clear, thank you for you answer.
          – Markus Steiner
          Jul 28 at 17:21












           

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