If a theory $T$ has a model that satisfies sentence $s$, does every model of $T$ satisfy $s$?

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The question is straightforward as expressed in the title: If a theory $T$ has a model that satisfies sentence $s$, does every model of $T$ satisfy $s$? I think I know the answer as yes, but why this is the case, I cannot find or remember. What would be such proof?

logic model-theory

share|cite|improve this question

edited Jul 21 at 14:58

asked Jul 21 at 14:56

Mullock Brian

204

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I would have thought the answer was no. What is $s$ anyway?
  – Lord Shark the Unknown
  Jul 21 at 14:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown $s$ is a sentence.
  – Mullock Brian
  Jul 21 at 14:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is $T$ assumed to be complete?
  – Noah Schweber
  Jul 21 at 15:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $D_8$ is a model of the theory of groups and has 8 elements. Not every group has 8 elements.
  – jgon
  Jul 21 at 15:08
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

The question is straightforward as expressed in the title: If a theory $T$ has a model that satisfies sentence $s$, does every model of $T$ satisfy $s$? I think I know the answer as yes, but why this is the case, I cannot find or remember. What would be such proof?

logic model-theory

share|cite|improve this question

edited Jul 21 at 14:58

asked Jul 21 at 14:56

Mullock Brian

204

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I would have thought the answer was no. What is $s$ anyway?
  – Lord Shark the Unknown
  Jul 21 at 14:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown $s$ is a sentence.
  – Mullock Brian
  Jul 21 at 14:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is $T$ assumed to be complete?
  – Noah Schweber
  Jul 21 at 15:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $D_8$ is a model of the theory of groups and has 8 elements. Not every group has 8 elements.
  – jgon
  Jul 21 at 15:08
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

The question is straightforward as expressed in the title: If a theory $T$ has a model that satisfies sentence $s$, does every model of $T$ satisfy $s$? I think I know the answer as yes, but why this is the case, I cannot find or remember. What would be such proof?

logic model-theory

share|cite|improve this question

edited Jul 21 at 14:58

asked Jul 21 at 14:56

Mullock Brian

204

The question is straightforward as expressed in the title: If a theory $T$ has a model that satisfies sentence $s$, does every model of $T$ satisfy $s$? I think I know the answer as yes, but why this is the case, I cannot find or remember. What would be such proof?

logic model-theory

share|cite|improve this question

edited Jul 21 at 14:58

asked Jul 21 at 14:56

Mullock Brian

204

share|cite|improve this question

share|cite|improve this question

edited Jul 21 at 14:58

edited Jul 21 at 14:58

edited Jul 21 at 14:58

asked Jul 21 at 14:56

Mullock Brian

204

asked Jul 21 at 14:56

Mullock Brian

204

asked Jul 21 at 14:56

Mullock Brian

204

204

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I would have thought the answer was no. What is $s$ anyway?
  – Lord Shark the Unknown
  Jul 21 at 14:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown $s$ is a sentence.
  – Mullock Brian
  Jul 21 at 14:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is $T$ assumed to be complete?
  – Noah Schweber
  Jul 21 at 15:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $D_8$ is a model of the theory of groups and has 8 elements. Not every group has 8 elements.
  – jgon
  Jul 21 at 15:08
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I would have thought the answer was no. What is $s$ anyway?
  – Lord Shark the Unknown
  Jul 21 at 14:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @LordSharktheUnknown $s$ is a sentence.
  – Mullock Brian
  Jul 21 at 14:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is $T$ assumed to be complete?
  – Noah Schweber
  Jul 21 at 15:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $D_8$ is a model of the theory of groups and has 8 elements. Not every group has 8 elements.
  – jgon
  Jul 21 at 15:08
  

  
  
  
  

2

2

I would have thought the answer was no. What is $s$ anyway?
– Lord Shark the Unknown
Jul 21 at 14:57

I would have thought the answer was no. What is $s$ anyway?
– Lord Shark the Unknown
Jul 21 at 14:57

@LordSharktheUnknown $s$ is a sentence.
– Mullock Brian
Jul 21 at 14:59

@LordSharktheUnknown $s$ is a sentence.
– Mullock Brian
Jul 21 at 14:59

1

1

Is $T$ assumed to be complete?
– Noah Schweber
Jul 21 at 15:05

Is $T$ assumed to be complete?
– Noah Schweber
Jul 21 at 15:05

2

2

$D_8$ is a model of the theory of groups and has 8 elements. Not every group has 8 elements.
– jgon
Jul 21 at 15:08

$D_8$ is a model of the theory of groups and has 8 elements. Not every group has 8 elements.
– jgon
Jul 21 at 15:08

add a comment | 

1 Answer
1

active

oldest

votes

up vote
4
down vote



accepted

The answer, as stated above, is no (unless we assume that $T$ is complete).

If you know a bit of abstract algebra, the simplest natural counterexample to your guess is probably to take $T$ to be the group axioms and $s$ to be the sentence "$forall xforall y(x*y=y*x)$." Some but not all groups are abelian, which is to say that some but not all models of $T$ satisfy $s$.

---

Leaving abstract algebra aside, here's a "purely combinatorial" counterexample:

Let $T$ be the empty theory, $T=emptyset$. Then every structure satisfies $T$, vacuously: for $mathcalM$ an arbitrary structure, we have "$mathcalMmodelsvarphi$ for every $varphiin T$" since there aren't any $varphiin T$ to begin with.

Now consider the sentence $s:=$ "$exists xexists y(xnot=y)$." The structures satisfying this sentence are exactlyt he structures with more than one element. Since there are structures which do have more than one element, and structures which don't have more than one element, that means there are models of $T$ which disagree about whether $s$ is true or false.

share|cite|improve this answer

answered Jul 21 at 16:25

Noah Schweber

111k9140261

• 
  
  
  

  
  

  
  
  
  
  
  

  
  so suppose $T$ is PA (Peano). PA is not complete, if I understand completeness as referring to a theory instead of logic. Then for PA, it is possible that there exists a model that satisfies $s$, while there exists a model that satisfies $neg s$. Is this correct understanding following from this answer?
  – Mullock Brian
  Jul 21 at 21:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MullockBrian Yes (I mean, it depends on what $s$ is of course). Completeness of a logic says "For every theory $T$ and every sentence $varphi$, if every model of $T$ satisfies $varphi$ then $Tvdashvarphi$." Note that completeness isn't really a property of a logic, but rather a property of a logic together with a proof system (the proof system is complete with respect to the logic). A theory is complete, by contrast, if for any sentence $varphi$ it either proves $varphi$ or proves $negvarphi$. (cont'd)
  – Noah Schweber
  Jul 21 at 21:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The completeness theorem (for first-order logic) says that a wide variety of proof systems are complete with respect to first-order logic. Incidentally, the completeness theorem (for a proof system $mathfrakp)$ is equivalent to the statement "A theory $T$ is complete (in the sense of $mathfrakp$) iff all its models are elementarily equivalent."
  – Noah Schweber
  Jul 21 at 21:09
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

The answer, as stated above, is no (unless we assume that $T$ is complete).

If you know a bit of abstract algebra, the simplest natural counterexample to your guess is probably to take $T$ to be the group axioms and $s$ to be the sentence "$forall xforall y(x*y=y*x)$." Some but not all groups are abelian, which is to say that some but not all models of $T$ satisfy $s$.

---

Leaving abstract algebra aside, here's a "purely combinatorial" counterexample:

Let $T$ be the empty theory, $T=emptyset$. Then every structure satisfies $T$, vacuously: for $mathcalM$ an arbitrary structure, we have "$mathcalMmodelsvarphi$ for every $varphiin T$" since there aren't any $varphiin T$ to begin with.

Now consider the sentence $s:=$ "$exists xexists y(xnot=y)$." The structures satisfying this sentence are exactlyt he structures with more than one element. Since there are structures which do have more than one element, and structures which don't have more than one element, that means there are models of $T$ which disagree about whether $s$ is true or false.

share|cite|improve this answer

answered Jul 21 at 16:25

Noah Schweber

111k9140261

• 
  
  
  

  
  

  
  
  
  
  
  

  
  so suppose $T$ is PA (Peano). PA is not complete, if I understand completeness as referring to a theory instead of logic. Then for PA, it is possible that there exists a model that satisfies $s$, while there exists a model that satisfies $neg s$. Is this correct understanding following from this answer?
  – Mullock Brian
  Jul 21 at 21:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MullockBrian Yes (I mean, it depends on what $s$ is of course). Completeness of a logic says "For every theory $T$ and every sentence $varphi$, if every model of $T$ satisfies $varphi$ then $Tvdashvarphi$." Note that completeness isn't really a property of a logic, but rather a property of a logic together with a proof system (the proof system is complete with respect to the logic). A theory is complete, by contrast, if for any sentence $varphi$ it either proves $varphi$ or proves $negvarphi$. (cont'd)
  – Noah Schweber
  Jul 21 at 21:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The completeness theorem (for first-order logic) says that a wide variety of proof systems are complete with respect to first-order logic. Incidentally, the completeness theorem (for a proof system $mathfrakp)$ is equivalent to the statement "A theory $T$ is complete (in the sense of $mathfrakp$) iff all its models are elementarily equivalent."
  – Noah Schweber
  Jul 21 at 21:09
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

The answer, as stated above, is no (unless we assume that $T$ is complete).

If you know a bit of abstract algebra, the simplest natural counterexample to your guess is probably to take $T$ to be the group axioms and $s$ to be the sentence "$forall xforall y(x*y=y*x)$." Some but not all groups are abelian, which is to say that some but not all models of $T$ satisfy $s$.

---

Leaving abstract algebra aside, here's a "purely combinatorial" counterexample:

Let $T$ be the empty theory, $T=emptyset$. Then every structure satisfies $T$, vacuously: for $mathcalM$ an arbitrary structure, we have "$mathcalMmodelsvarphi$ for every $varphiin T$" since there aren't any $varphiin T$ to begin with.

Now consider the sentence $s:=$ "$exists xexists y(xnot=y)$." The structures satisfying this sentence are exactlyt he structures with more than one element. Since there are structures which do have more than one element, and structures which don't have more than one element, that means there are models of $T$ which disagree about whether $s$ is true or false.

share|cite|improve this answer

answered Jul 21 at 16:25

Noah Schweber

111k9140261

• 
  
  
  

  
  

  
  
  
  
  
  

  
  so suppose $T$ is PA (Peano). PA is not complete, if I understand completeness as referring to a theory instead of logic. Then for PA, it is possible that there exists a model that satisfies $s$, while there exists a model that satisfies $neg s$. Is this correct understanding following from this answer?
  – Mullock Brian
  Jul 21 at 21:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MullockBrian Yes (I mean, it depends on what $s$ is of course). Completeness of a logic says "For every theory $T$ and every sentence $varphi$, if every model of $T$ satisfies $varphi$ then $Tvdashvarphi$." Note that completeness isn't really a property of a logic, but rather a property of a logic together with a proof system (the proof system is complete with respect to the logic). A theory is complete, by contrast, if for any sentence $varphi$ it either proves $varphi$ or proves $negvarphi$. (cont'd)
  – Noah Schweber
  Jul 21 at 21:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The completeness theorem (for first-order logic) says that a wide variety of proof systems are complete with respect to first-order logic. Incidentally, the completeness theorem (for a proof system $mathfrakp)$ is equivalent to the statement "A theory $T$ is complete (in the sense of $mathfrakp$) iff all its models are elementarily equivalent."
  – Noah Schweber
  Jul 21 at 21:09
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

The answer, as stated above, is no (unless we assume that $T$ is complete).

If you know a bit of abstract algebra, the simplest natural counterexample to your guess is probably to take $T$ to be the group axioms and $s$ to be the sentence "$forall xforall y(x*y=y*x)$." Some but not all groups are abelian, which is to say that some but not all models of $T$ satisfy $s$.

---

Leaving abstract algebra aside, here's a "purely combinatorial" counterexample:

Let $T$ be the empty theory, $T=emptyset$. Then every structure satisfies $T$, vacuously: for $mathcalM$ an arbitrary structure, we have "$mathcalMmodelsvarphi$ for every $varphiin T$" since there aren't any $varphiin T$ to begin with.

Now consider the sentence $s:=$ "$exists xexists y(xnot=y)$." The structures satisfying this sentence are exactlyt he structures with more than one element. Since there are structures which do have more than one element, and structures which don't have more than one element, that means there are models of $T$ which disagree about whether $s$ is true or false.

share|cite|improve this answer

answered Jul 21 at 16:25

Noah Schweber

111k9140261

The answer, as stated above, is no (unless we assume that $T$ is complete).

If you know a bit of abstract algebra, the simplest natural counterexample to your guess is probably to take $T$ to be the group axioms and $s$ to be the sentence "$forall xforall y(x*y=y*x)$." Some but not all groups are abelian, which is to say that some but not all models of $T$ satisfy $s$.

---

Leaving abstract algebra aside, here's a "purely combinatorial" counterexample:

Let $T$ be the empty theory, $T=emptyset$. Then every structure satisfies $T$, vacuously: for $mathcalM$ an arbitrary structure, we have "$mathcalMmodelsvarphi$ for every $varphiin T$" since there aren't any $varphiin T$ to begin with.

Now consider the sentence $s:=$ "$exists xexists y(xnot=y)$." The structures satisfying this sentence are exactlyt he structures with more than one element. Since there are structures which do have more than one element, and structures which don't have more than one element, that means there are models of $T$ which disagree about whether $s$ is true or false.

share|cite|improve this answer

answered Jul 21 at 16:25

Noah Schweber

111k9140261

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 16:25

Noah Schweber

111k9140261

answered Jul 21 at 16:25

Noah Schweber

111k9140261

answered Jul 21 at 16:25

Noah Schweber

111k9140261

111k9140261

• 
  
  
  

  
  

  
  
  
  
  
  

  
  so suppose $T$ is PA (Peano). PA is not complete, if I understand completeness as referring to a theory instead of logic. Then for PA, it is possible that there exists a model that satisfies $s$, while there exists a model that satisfies $neg s$. Is this correct understanding following from this answer?
  – Mullock Brian
  Jul 21 at 21:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MullockBrian Yes (I mean, it depends on what $s$ is of course). Completeness of a logic says "For every theory $T$ and every sentence $varphi$, if every model of $T$ satisfies $varphi$ then $Tvdashvarphi$." Note that completeness isn't really a property of a logic, but rather a property of a logic together with a proof system (the proof system is complete with respect to the logic). A theory is complete, by contrast, if for any sentence $varphi$ it either proves $varphi$ or proves $negvarphi$. (cont'd)
  – Noah Schweber
  Jul 21 at 21:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The completeness theorem (for first-order logic) says that a wide variety of proof systems are complete with respect to first-order logic. Incidentally, the completeness theorem (for a proof system $mathfrakp)$ is equivalent to the statement "A theory $T$ is complete (in the sense of $mathfrakp$) iff all its models are elementarily equivalent."
  – Noah Schweber
  Jul 21 at 21:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  so suppose $T$ is PA (Peano). PA is not complete, if I understand completeness as referring to a theory instead of logic. Then for PA, it is possible that there exists a model that satisfies $s$, while there exists a model that satisfies $neg s$. Is this correct understanding following from this answer?
  – Mullock Brian
  Jul 21 at 21:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MullockBrian Yes (I mean, it depends on what $s$ is of course). Completeness of a logic says "For every theory $T$ and every sentence $varphi$, if every model of $T$ satisfies $varphi$ then $Tvdashvarphi$." Note that completeness isn't really a property of a logic, but rather a property of a logic together with a proof system (the proof system is complete with respect to the logic). A theory is complete, by contrast, if for any sentence $varphi$ it either proves $varphi$ or proves $negvarphi$. (cont'd)
  – Noah Schweber
  Jul 21 at 21:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The completeness theorem (for first-order logic) says that a wide variety of proof systems are complete with respect to first-order logic. Incidentally, the completeness theorem (for a proof system $mathfrakp)$ is equivalent to the statement "A theory $T$ is complete (in the sense of $mathfrakp$) iff all its models are elementarily equivalent."
  – Noah Schweber
  Jul 21 at 21:09
  

  
  
  
  

so suppose $T$ is PA (Peano). PA is not complete, if I understand completeness as referring to a theory instead of logic. Then for PA, it is possible that there exists a model that satisfies $s$, while there exists a model that satisfies $neg s$. Is this correct understanding following from this answer?
– Mullock Brian
Jul 21 at 21:03

so suppose $T$ is PA (Peano). PA is not complete, if I understand completeness as referring to a theory instead of logic. Then for PA, it is possible that there exists a model that satisfies $s$, while there exists a model that satisfies $neg s$. Is this correct understanding following from this answer?
– Mullock Brian
Jul 21 at 21:03

@MullockBrian Yes (I mean, it depends on what $s$ is of course). Completeness of a logic says "For every theory $T$ and every sentence $varphi$, if every model of $T$ satisfies $varphi$ then $Tvdashvarphi$." Note that completeness isn't really a property of a logic, but rather a property of a logic together with a proof system (the proof system is complete with respect to the logic). A theory is complete, by contrast, if for any sentence $varphi$ it either proves $varphi$ or proves $negvarphi$. (cont'd)
– Noah Schweber
Jul 21 at 21:07

@MullockBrian Yes (I mean, it depends on what $s$ is of course). Completeness of a logic says "For every theory $T$ and every sentence $varphi$, if every model of $T$ satisfies $varphi$ then $Tvdashvarphi$." Note that completeness isn't really a property of a logic, but rather a property of a logic together with a proof system (the proof system is complete with respect to the logic). A theory is complete, by contrast, if for any sentence $varphi$ it either proves $varphi$ or proves $negvarphi$. (cont'd)
– Noah Schweber
Jul 21 at 21:07

The completeness theorem (for first-order logic) says that a wide variety of proof systems are complete with respect to first-order logic. Incidentally, the completeness theorem (for a proof system $mathfrakp)$ is equivalent to the statement "A theory $T$ is complete (in the sense of $mathfrakp$) iff all its models are elementarily equivalent."
– Noah Schweber
Jul 21 at 21:09

The completeness theorem (for first-order logic) says that a wide variety of proof systems are complete with respect to first-order logic. Incidentally, the completeness theorem (for a proof system $mathfrakp)$ is equivalent to the statement "A theory $T$ is complete (in the sense of $mathfrakp$) iff all its models are elementarily equivalent."
– Noah Schweber
Jul 21 at 21:09

add a comment | 

 

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