Mixing
Proof that there exist no finite axiomatic system with Compactness Theorem
$begingroup$
Say we would like to prove that the class of all infinite groups $(G, circ, e)$ is not finitely axiomatizable by making use of the compactness theorem. We normally prove this by contradiction since we know the axiomatic system is defined by the union of $Phi_{group}$ (the standard FO axiomatic system for groups) and $Phi_{infty}$ (in FO logic, there exist $x_1$ to $x_n$ and they are all pairwise non-identical and that for all $n in mathbb{N}$).
Assuming there exist a $Phi_{infGrp} subseteq FO(tau)$ that finitely axiomatizes this class. Therefore, we can find a $psi in FO(tau)$ that axiomatizes this class too, since the axiomatic system is finite.
This implies that $Phi_{infGrp} cup {negpsi}$ is not satisfiable. We then go on to find a arbitrary, finite subset $Phi_0 subseteq Phi_{infGrp} cup {negpsi}$, and show that we can find a structure that is a model of this subset by making the said structure to have more elements than the formula with a largest $m in mathbb{N}$ in $Phi_0$. We then finally say that since structure $mathbb{A} models Phi_0$, by virtue of compactness theory, $Phi_{infGrp} cup {negpsi}$ is satsifiable.
The question is... how can $Phi_{infGrp} cup {negpsi}$ be satisfiable when $negpsi$ contains statements that negate the properties of a group. Wouldn't every structure (that is a group) not be a model for $Phi_{infGrp} cup {negpsi}$?
logic first-order-logic predicate-logic model-theory
edited Jan 26 at 21:06
Taroccoesbrocco
5,64271840
asked Aug 12 '14 at 11:46
mercurialmercurial
5862815
$endgroup$
add a comment |
$begingroup$
Say we would like to prove that the class of all infinite groups $(G, circ, e)$ is not finitely axiomatizable by making use of the compactness theorem. We normally prove this by contradiction since we know the axiomatic system is defined by the union of $Phi_{group}$ (the standard FO axiomatic system for groups) and $Phi_{infty}$ (in FO logic, there exist $x_1$ to $x_n$ and they are all pairwise non-identical and that for all $n in mathbb{N}$).
Assuming there exist a $Phi_{infGrp} subseteq FO(tau)$ that finitely axiomatizes this class. Therefore, we can find a $psi in FO(tau)$ that axiomatizes this class too, since the axiomatic system is finite.
This implies that $Phi_{infGrp} cup {negpsi}$ is not satisfiable. We then go on to find a arbitrary, finite subset $Phi_0 subseteq Phi_{infGrp} cup {negpsi}$, and show that we can find a structure that is a model of this subset by making the said structure to have more elements than the formula with a largest $m in mathbb{N}$ in $Phi_0$. We then finally say that since structure $mathbb{A} models Phi_0$, by virtue of compactness theory, $Phi_{infGrp} cup {negpsi}$ is satsifiable.
The question is... how can $Phi_{infGrp} cup {negpsi}$ be satisfiable when $negpsi$ contains statements that negate the properties of a group. Wouldn't every structure (that is a group) not be a model for $Phi_{infGrp} cup {negpsi}$?
logic first-order-logic predicate-logic model-theory
edited Jan 26 at 21:06
Taroccoesbrocco
5,64271840
asked Aug 12 '14 at 11:46
mercurialmercurial
5862815
$endgroup$
add a comment |
$begingroup$
Say we would like to prove that the class of all infinite groups $(G, circ, e)$ is not finitely axiomatizable by making use of the compactness theorem. We normally prove this by contradiction since we know the axiomatic system is defined by the union of $Phi_{group}$ (the standard FO axiomatic system for groups) and $Phi_{infty}$ (in FO logic, there exist $x_1$ to $x_n$ and they are all pairwise non-identical and that for all $n in mathbb{N}$).
Assuming there exist a $Phi_{infGrp} subseteq FO(tau)$ that finitely axiomatizes this class. Therefore, we can find a $psi in FO(tau)$ that axiomatizes this class too, since the axiomatic system is finite.
This implies that $Phi_{infGrp} cup {negpsi}$ is not satisfiable. We then go on to find a arbitrary, finite subset $Phi_0 subseteq Phi_{infGrp} cup {negpsi}$, and show that we can find a structure that is a model of this subset by making the said structure to have more elements than the formula with a largest $m in mathbb{N}$ in $Phi_0$. We then finally say that since structure $mathbb{A} models Phi_0$, by virtue of compactness theory, $Phi_{infGrp} cup {negpsi}$ is satsifiable.
The question is... how can $Phi_{infGrp} cup {negpsi}$ be satisfiable when $negpsi$ contains statements that negate the properties of a group. Wouldn't every structure (that is a group) not be a model for $Phi_{infGrp} cup {negpsi}$?
logic first-order-logic predicate-logic model-theory
edited Jan 26 at 21:06
Taroccoesbrocco
5,64271840
asked Aug 12 '14 at 11:46
mercurialmercurial
5862815
$endgroup$
Say we would like to prove that the class of all infinite groups $(G, circ, e)$ is not finitely axiomatizable by making use of the compactness theorem. We normally prove this by contradiction since we know the axiomatic system is defined by the union of $Phi_{group}$ (the standard FO axiomatic system for groups) and $Phi_{infty}$ (in FO logic, there exist $x_1$ to $x_n$ and they are all pairwise non-identical and that for all $n in mathbb{N}$).
Assuming there exist a $Phi_{infGrp} subseteq FO(tau)$ that finitely axiomatizes this class. Therefore, we can find a $psi in FO(tau)$ that axiomatizes this class too, since the axiomatic system is finite.
This implies that $Phi_{infGrp} cup {negpsi}$ is not satisfiable. We then go on to find a arbitrary, finite subset $Phi_0 subseteq Phi_{infGrp} cup {negpsi}$, and show that we can find a structure that is a model of this subset by making the said structure to have more elements than the formula with a largest $m in mathbb{N}$ in $Phi_0$. We then finally say that since structure $mathbb{A} models Phi_0$, by virtue of compactness theory, $Phi_{infGrp} cup {negpsi}$ is satsifiable.
The question is... how can $Phi_{infGrp} cup {negpsi}$ be satisfiable when $negpsi$ contains statements that negate the properties of a group. Wouldn't every structure (that is a group) not be a model for $Phi_{infGrp} cup {negpsi}$?
logic first-order-logic predicate-logic model-theory
logic first-order-logic predicate-logic model-theory
edited Jan 26 at 21:06
Taroccoesbrocco
5,64271840
asked Aug 12 '14 at 11:46
mercurialmercurial
5862815
edited Jan 26 at 21:06
Taroccoesbrocco
5,64271840
asked Aug 12 '14 at 11:46
mercurialmercurial
5862815
edited Jan 26 at 21:06
Taroccoesbrocco
5,64271840
edited Jan 26 at 21:06
Taroccoesbrocco
5,64271840
edited Jan 26 at 21:06
Taroccoesbrocco
5,64271840
5,64271840
asked Aug 12 '14 at 11:46
mercurialmercurial
5862815
asked Aug 12 '14 at 11:46
mercurialmercurial
5862815
asked Aug 12 '14 at 11:46
mercurialmercurial
5862815
5862815
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is nothing in $lnotpsi$ that needs to negate the properties of a group. Assume finite axiomatizability. So we can take the axioms to be $alpha_1,dots,alpha_m$, the usual axioms of group theory, plus additional special axioms $beta_1,dots,beta_n$, which in conjunction with the group axioms force the group to be infinite. Then
$psi$ is the sentence
$$alpha_1landcdotsland alpha_m land beta_1landdotsland beta_n.$$
The sentence $lnot psi$ is then the sentence
$$lnotalpha_1lorcdotslor lnotalpha_m lor lnotbeta_1lordotslor lnotbeta_n.$$
This is a disjunction. No group property needs to fail.
answered Aug 12 '14 at 12:02
André NicolasAndré Nicolas
454k36432819
$endgroup$
$begingroup$
Ah yea true that didn't occur to me! Thanks!
$endgroup$
– mercurial
Aug 12 '14 at 12:25
$begingroup$
You are welcome. It is natural to think that a group-theoretic axiom is being negated.
$endgroup$
– André Nicolas
Aug 12 '14 at 13:13
add a comment |
$begingroup$
The following may be a more intuitive proof:
Let $mathcal{L}$ be the first order language of groups. Let $K$ denote the class of infinite groups.
If $K$ was finitely axiomatizable, then by taking conjunctions, one may assume that there is a single formula $psi$ such that $K$ is the set of the all $mathcal{L}$-structures $M$ such that $M models psi$. Since the group axioms are finite, let $Phi$ be the conjunction of all the group axioms. Then $ Phi wedge neg psi$ axiomatizes the class of all finite groups. Let $mathcal{L}' = mathcal{L} cup {c_i : i < omega}$. $mathcal{L}'$ is the expanded language with infinitely many new constant symbols. Let $T = {Phi wedge neg psi} cup {c_i neq c_j : i neq j}$. If $Delta subseteq T$ is a finite set, then only finitely many new constant symbols $c_{i_0}, ..., c_{i_n}$ were mentioned in any sentence of $Delta$. Take any group $G$ of size $n$. Make $G$ into a $mathcal{L}'$-structure by interpreting $c_{i_0}, ..., c_{i_n}$ as $n$ distinct elements of $G$. (You can interpret the other constants to be whatever you want.) Then $G$ as a $mathcal{L}'$-structure satisfies $Delta$. So $Delta$ is satisfiable. By the compactness theorem, there exists a $G$ such that $G models T$. Clearly $G$ must be an infinite $mathcal{L}'$-structure which satisfies $Phi wedge neg psi$. The reduct of $G$ back to $mathcal{L}$ is then an infinite group satisfying $Phi wedge neg psi$. This is a contradiction since $Phi wedge neg psi$ was suppose to be an axiomatization of finite groups.
answered Aug 12 '14 at 12:08
WilliamWilliam
17.3k22256
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
There is nothing in $lnotpsi$ that needs to negate the properties of a group. Assume finite axiomatizability. So we can take the axioms to be $alpha_1,dots,alpha_m$, the usual axioms of group theory, plus additional special axioms $beta_1,dots,beta_n$, which in conjunction with the group axioms force the group to be infinite. Then
$psi$ is the sentence
$$alpha_1landcdotsland alpha_m land beta_1landdotsland beta_n.$$
The sentence $lnot psi$ is then the sentence
$$lnotalpha_1lorcdotslor lnotalpha_m lor lnotbeta_1lordotslor lnotbeta_n.$$
This is a disjunction. No group property needs to fail.
answered Aug 12 '14 at 12:02
André NicolasAndré Nicolas
454k36432819
$endgroup$
$begingroup$
Ah yea true that didn't occur to me! Thanks!
$endgroup$
– mercurial
Aug 12 '14 at 12:25
$begingroup$
You are welcome. It is natural to think that a group-theoretic axiom is being negated.
$endgroup$
– André Nicolas
Aug 12 '14 at 13:13
add a comment |
$begingroup$
There is nothing in $lnotpsi$ that needs to negate the properties of a group. Assume finite axiomatizability. So we can take the axioms to be $alpha_1,dots,alpha_m$, the usual axioms of group theory, plus additional special axioms $beta_1,dots,beta_n$, which in conjunction with the group axioms force the group to be infinite. Then
$psi$ is the sentence
$$alpha_1landcdotsland alpha_m land beta_1landdotsland beta_n.$$
The sentence $lnot psi$ is then the sentence
$$lnotalpha_1lorcdotslor lnotalpha_m lor lnotbeta_1lordotslor lnotbeta_n.$$
This is a disjunction. No group property needs to fail.
answered Aug 12 '14 at 12:02
André NicolasAndré Nicolas
454k36432819
$endgroup$
$begingroup$
Ah yea true that didn't occur to me! Thanks!
$endgroup$
– mercurial
Aug 12 '14 at 12:25
$begingroup$
You are welcome. It is natural to think that a group-theoretic axiom is being negated.
$endgroup$
– André Nicolas
Aug 12 '14 at 13:13
add a comment |
$begingroup$
There is nothing in $lnotpsi$ that needs to negate the properties of a group. Assume finite axiomatizability. So we can take the axioms to be $alpha_1,dots,alpha_m$, the usual axioms of group theory, plus additional special axioms $beta_1,dots,beta_n$, which in conjunction with the group axioms force the group to be infinite. Then
$psi$ is the sentence
$$alpha_1landcdotsland alpha_m land beta_1landdotsland beta_n.$$
The sentence $lnot psi$ is then the sentence
$$lnotalpha_1lorcdotslor lnotalpha_m lor lnotbeta_1lordotslor lnotbeta_n.$$
This is a disjunction. No group property needs to fail.
answered Aug 12 '14 at 12:02
André NicolasAndré Nicolas
454k36432819
$endgroup$
There is nothing in $lnotpsi$ that needs to negate the properties of a group. Assume finite axiomatizability. So we can take the axioms to be $alpha_1,dots,alpha_m$, the usual axioms of group theory, plus additional special axioms $beta_1,dots,beta_n$, which in conjunction with the group axioms force the group to be infinite. Then
$psi$ is the sentence
$$alpha_1landcdotsland alpha_m land beta_1landdotsland beta_n.$$
The sentence $lnot psi$ is then the sentence
$$lnotalpha_1lorcdotslor lnotalpha_m lor lnotbeta_1lordotslor lnotbeta_n.$$
This is a disjunction. No group property needs to fail.
answered Aug 12 '14 at 12:02
André NicolasAndré Nicolas
454k36432819
answered Aug 12 '14 at 12:02
André NicolasAndré Nicolas
454k36432819
answered Aug 12 '14 at 12:02
André NicolasAndré Nicolas
454k36432819
answered Aug 12 '14 at 12:02
André NicolasAndré Nicolas
454k36432819
454k36432819
$begingroup$
Ah yea true that didn't occur to me! Thanks!
$endgroup$
– mercurial
Aug 12 '14 at 12:25
$begingroup$
You are welcome. It is natural to think that a group-theoretic axiom is being negated.
$endgroup$
– André Nicolas
Aug 12 '14 at 13:13
add a comment |
$begingroup$
Ah yea true that didn't occur to me! Thanks!
$endgroup$
– mercurial
Aug 12 '14 at 12:25
$begingroup$
You are welcome. It is natural to think that a group-theoretic axiom is being negated.
$endgroup$
– André Nicolas
Aug 12 '14 at 13:13
$begingroup$
Ah yea true that didn't occur to me! Thanks!
$endgroup$
– mercurial
Aug 12 '14 at 12:25
$begingroup$
Ah yea true that didn't occur to me! Thanks!
$endgroup$
– mercurial
Aug 12 '14 at 12:25
$begingroup$
You are welcome. It is natural to think that a group-theoretic axiom is being negated.
$endgroup$
– André Nicolas
Aug 12 '14 at 13:13
$begingroup$
You are welcome. It is natural to think that a group-theoretic axiom is being negated.
$endgroup$
– André Nicolas
Aug 12 '14 at 13:13
add a comment |
$begingroup$
The following may be a more intuitive proof:
Let $mathcal{L}$ be the first order language of groups. Let $K$ denote the class of infinite groups.
If $K$ was finitely axiomatizable, then by taking conjunctions, one may assume that there is a single formula $psi$ such that $K$ is the set of the all $mathcal{L}$-structures $M$ such that $M models psi$. Since the group axioms are finite, let $Phi$ be the conjunction of all the group axioms. Then $ Phi wedge neg psi$ axiomatizes the class of all finite groups. Let $mathcal{L}' = mathcal{L} cup {c_i : i < omega}$. $mathcal{L}'$ is the expanded language with infinitely many new constant symbols. Let $T = {Phi wedge neg psi} cup {c_i neq c_j : i neq j}$. If $Delta subseteq T$ is a finite set, then only finitely many new constant symbols $c_{i_0}, ..., c_{i_n}$ were mentioned in any sentence of $Delta$. Take any group $G$ of size $n$. Make $G$ into a $mathcal{L}'$-structure by interpreting $c_{i_0}, ..., c_{i_n}$ as $n$ distinct elements of $G$. (You can interpret the other constants to be whatever you want.) Then $G$ as a $mathcal{L}'$-structure satisfies $Delta$. So $Delta$ is satisfiable. By the compactness theorem, there exists a $G$ such that $G models T$. Clearly $G$ must be an infinite $mathcal{L}'$-structure which satisfies $Phi wedge neg psi$. The reduct of $G$ back to $mathcal{L}$ is then an infinite group satisfying $Phi wedge neg psi$. This is a contradiction since $Phi wedge neg psi$ was suppose to be an axiomatization of finite groups.
answered Aug 12 '14 at 12:08
WilliamWilliam
17.3k22256
$endgroup$
add a comment |
$begingroup$
The following may be a more intuitive proof:
Let $mathcal{L}$ be the first order language of groups. Let $K$ denote the class of infinite groups.
If $K$ was finitely axiomatizable, then by taking conjunctions, one may assume that there is a single formula $psi$ such that $K$ is the set of the all $mathcal{L}$-structures $M$ such that $M models psi$. Since the group axioms are finite, let $Phi$ be the conjunction of all the group axioms. Then $ Phi wedge neg psi$ axiomatizes the class of all finite groups. Let $mathcal{L}' = mathcal{L} cup {c_i : i < omega}$. $mathcal{L}'$ is the expanded language with infinitely many new constant symbols. Let $T = {Phi wedge neg psi} cup {c_i neq c_j : i neq j}$. If $Delta subseteq T$ is a finite set, then only finitely many new constant symbols $c_{i_0}, ..., c_{i_n}$ were mentioned in any sentence of $Delta$. Take any group $G$ of size $n$. Make $G$ into a $mathcal{L}'$-structure by interpreting $c_{i_0}, ..., c_{i_n}$ as $n$ distinct elements of $G$. (You can interpret the other constants to be whatever you want.) Then $G$ as a $mathcal{L}'$-structure satisfies $Delta$. So $Delta$ is satisfiable. By the compactness theorem, there exists a $G$ such that $G models T$. Clearly $G$ must be an infinite $mathcal{L}'$-structure which satisfies $Phi wedge neg psi$. The reduct of $G$ back to $mathcal{L}$ is then an infinite group satisfying $Phi wedge neg psi$. This is a contradiction since $Phi wedge neg psi$ was suppose to be an axiomatization of finite groups.
answered Aug 12 '14 at 12:08
WilliamWilliam
17.3k22256
$endgroup$
add a comment |
$begingroup$
The following may be a more intuitive proof:
Let $mathcal{L}$ be the first order language of groups. Let $K$ denote the class of infinite groups.
If $K$ was finitely axiomatizable, then by taking conjunctions, one may assume that there is a single formula $psi$ such that $K$ is the set of the all $mathcal{L}$-structures $M$ such that $M models psi$. Since the group axioms are finite, let $Phi$ be the conjunction of all the group axioms. Then $ Phi wedge neg psi$ axiomatizes the class of all finite groups. Let $mathcal{L}' = mathcal{L} cup {c_i : i < omega}$. $mathcal{L}'$ is the expanded language with infinitely many new constant symbols. Let $T = {Phi wedge neg psi} cup {c_i neq c_j : i neq j}$. If $Delta subseteq T$ is a finite set, then only finitely many new constant symbols $c_{i_0}, ..., c_{i_n}$ were mentioned in any sentence of $Delta$. Take any group $G$ of size $n$. Make $G$ into a $mathcal{L}'$-structure by interpreting $c_{i_0}, ..., c_{i_n}$ as $n$ distinct elements of $G$. (You can interpret the other constants to be whatever you want.) Then $G$ as a $mathcal{L}'$-structure satisfies $Delta$. So $Delta$ is satisfiable. By the compactness theorem, there exists a $G$ such that $G models T$. Clearly $G$ must be an infinite $mathcal{L}'$-structure which satisfies $Phi wedge neg psi$. The reduct of $G$ back to $mathcal{L}$ is then an infinite group satisfying $Phi wedge neg psi$. This is a contradiction since $Phi wedge neg psi$ was suppose to be an axiomatization of finite groups.
answered Aug 12 '14 at 12:08
WilliamWilliam
17.3k22256
$endgroup$
The following may be a more intuitive proof:
Let $mathcal{L}$ be the first order language of groups. Let $K$ denote the class of infinite groups.
If $K$ was finitely axiomatizable, then by taking conjunctions, one may assume that there is a single formula $psi$ such that $K$ is the set of the all $mathcal{L}$-structures $M$ such that $M models psi$. Since the group axioms are finite, let $Phi$ be the conjunction of all the group axioms. Then $ Phi wedge neg psi$ axiomatizes the class of all finite groups. Let $mathcal{L}' = mathcal{L} cup {c_i : i < omega}$. $mathcal{L}'$ is the expanded language with infinitely many new constant symbols. Let $T = {Phi wedge neg psi} cup {c_i neq c_j : i neq j}$. If $Delta subseteq T$ is a finite set, then only finitely many new constant symbols $c_{i_0}, ..., c_{i_n}$ were mentioned in any sentence of $Delta$. Take any group $G$ of size $n$. Make $G$ into a $mathcal{L}'$-structure by interpreting $c_{i_0}, ..., c_{i_n}$ as $n$ distinct elements of $G$. (You can interpret the other constants to be whatever you want.) Then $G$ as a $mathcal{L}'$-structure satisfies $Delta$. So $Delta$ is satisfiable. By the compactness theorem, there exists a $G$ such that $G models T$. Clearly $G$ must be an infinite $mathcal{L}'$-structure which satisfies $Phi wedge neg psi$. The reduct of $G$ back to $mathcal{L}$ is then an infinite group satisfying $Phi wedge neg psi$. This is a contradiction since $Phi wedge neg psi$ was suppose to be an axiomatization of finite groups.
answered Aug 12 '14 at 12:08
WilliamWilliam
17.3k22256
edited Aug 12 '14 at 12:17
answered Aug 12 '14 at 12:08
WilliamWilliam
17.3k22256
answered Aug 12 '14 at 12:08
WilliamWilliam
17.3k22256
answered Aug 12 '14 at 12:08
WilliamWilliam
17.3k22256
17.3k22256
add a comment |
add a comment |
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