$T models phi(c_1,ldots,c_n)$ implies $T models forall x_1,ldots,x_n phi(x_1,ldots,x_n)$?
Let $L$ be a language and $T$ be an $L$-theory.
Let $L'=L cup C$ where $C$ is a set of new constant symbols.
Suppose $phi(x_1,ldots,x_n)$ is an $L$-formula and $(c_1,ldots,c_n) in C^n$.
(1) Prove that if $T models phi(c_1,ldots,c_n)$ in $L'$ then $T models forall x_1,ldots,x_n phi(x_1,ldots,x_n)$ in $L$.
How can this statement be true? Shouldn't it be $T models phi(c_1,ldots,c_n)$ in $L'$ implies $T models exists x_1,ldots,x_n phi(x_1,ldots,x_n)$ in $L$?
Moreover, prove that $T$ admits quantifier elimination as an $L'$-theory if and only if $T$ admits quantifier elimination as an $L$-theory.
I proved the $(Rightarrow)$ part by assuming (1) is true. So is (1) really true? And how can I prove the converse?
first-order-logic model-theory
asked Nov 28 '18 at 12:24
bbw
47038
add a comment |
Let $L$ be a language and $T$ be an $L$-theory.
Let $L'=L cup C$ where $C$ is a set of new constant symbols.
Suppose $phi(x_1,ldots,x_n)$ is an $L$-formula and $(c_1,ldots,c_n) in C^n$.
(1) Prove that if $T models phi(c_1,ldots,c_n)$ in $L'$ then $T models forall x_1,ldots,x_n phi(x_1,ldots,x_n)$ in $L$.
How can this statement be true? Shouldn't it be $T models phi(c_1,ldots,c_n)$ in $L'$ implies $T models exists x_1,ldots,x_n phi(x_1,ldots,x_n)$ in $L$?
Moreover, prove that $T$ admits quantifier elimination as an $L'$-theory if and only if $T$ admits quantifier elimination as an $L$-theory.
I proved the $(Rightarrow)$ part by assuming (1) is true. So is (1) really true? And how can I prove the converse?
first-order-logic model-theory
asked Nov 28 '18 at 12:24
bbw
47038
A theory is a set of sentences; thus, $T vDash phi$ means that sentence $phi$ is a logical cons of $T$. If $T$ ia an $L$-theory, a model $mathcal M$ for $T$ does not specifiy how to interpret the new constants $c_i$. Thus, $TvDash phi$ means that every model $mathcal M$ of $T$ satisfies also $phi$, whatever the values of the domain $M$ of $mathcal M$ are assigned to the new constants $c_i$.
– Mauro ALLEGRANZA
Nov 28 '18 at 12:38
add a comment |
Let $L$ be a language and $T$ be an $L$-theory.
Let $L'=L cup C$ where $C$ is a set of new constant symbols.
Suppose $phi(x_1,ldots,x_n)$ is an $L$-formula and $(c_1,ldots,c_n) in C^n$.
(1) Prove that if $T models phi(c_1,ldots,c_n)$ in $L'$ then $T models forall x_1,ldots,x_n phi(x_1,ldots,x_n)$ in $L$.
How can this statement be true? Shouldn't it be $T models phi(c_1,ldots,c_n)$ in $L'$ implies $T models exists x_1,ldots,x_n phi(x_1,ldots,x_n)$ in $L$?
Moreover, prove that $T$ admits quantifier elimination as an $L'$-theory if and only if $T$ admits quantifier elimination as an $L$-theory.
I proved the $(Rightarrow)$ part by assuming (1) is true. So is (1) really true? And how can I prove the converse?
first-order-logic model-theory
asked Nov 28 '18 at 12:24
bbw
47038
Let $L$ be a language and $T$ be an $L$-theory.
Let $L'=L cup C$ where $C$ is a set of new constant symbols.
Suppose $phi(x_1,ldots,x_n)$ is an $L$-formula and $(c_1,ldots,c_n) in C^n$.
(1) Prove that if $T models phi(c_1,ldots,c_n)$ in $L'$ then $T models forall x_1,ldots,x_n phi(x_1,ldots,x_n)$ in $L$.
How can this statement be true? Shouldn't it be $T models phi(c_1,ldots,c_n)$ in $L'$ implies $T models exists x_1,ldots,x_n phi(x_1,ldots,x_n)$ in $L$?
Moreover, prove that $T$ admits quantifier elimination as an $L'$-theory if and only if $T$ admits quantifier elimination as an $L$-theory.
I proved the $(Rightarrow)$ part by assuming (1) is true. So is (1) really true? And how can I prove the converse?
first-order-logic model-theory
first-order-logic model-theory
asked Nov 28 '18 at 12:24
bbw
47038
asked Nov 28 '18 at 12:24
bbw
47038
asked Nov 28 '18 at 12:24
bbw
47038
asked Nov 28 '18 at 12:24
bbw
47038
asked Nov 28 '18 at 12:24
bbw
47038
47038
A theory is a set of sentences; thus, $T vDash phi$ means that sentence $phi$ is a logical cons of $T$. If $T$ ia an $L$-theory, a model $mathcal M$ for $T$ does not specifiy how to interpret the new constants $c_i$. Thus, $TvDash phi$ means that every model $mathcal M$ of $T$ satisfies also $phi$, whatever the values of the domain $M$ of $mathcal M$ are assigned to the new constants $c_i$.
– Mauro ALLEGRANZA
Nov 28 '18 at 12:38
add a comment |
A theory is a set of sentences; thus, $T vDash phi$ means that sentence $phi$ is a logical cons of $T$. If $T$ ia an $L$-theory, a model $mathcal M$ for $T$ does not specifiy how to interpret the new constants $c_i$. Thus, $TvDash phi$ means that every model $mathcal M$ of $T$ satisfies also $phi$, whatever the values of the domain $M$ of $mathcal M$ are assigned to the new constants $c_i$.
– Mauro ALLEGRANZA
Nov 28 '18 at 12:38
A theory is a set of sentences; thus, $T vDash phi$ means that sentence $phi$ is a logical cons of $T$. If $T$ ia an $L$-theory, a model $mathcal M$ for $T$ does not specifiy how to interpret the new constants $c_i$. Thus, $TvDash phi$ means that every model $mathcal M$ of $T$ satisfies also $phi$, whatever the values of the domain $M$ of $mathcal M$ are assigned to the new constants $c_i$.
– Mauro ALLEGRANZA
Nov 28 '18 at 12:38
A theory is a set of sentences; thus, $T vDash phi$ means that sentence $phi$ is a logical cons of $T$. If $T$ ia an $L$-theory, a model $mathcal M$ for $T$ does not specifiy how to interpret the new constants $c_i$. Thus, $TvDash phi$ means that every model $mathcal M$ of $T$ satisfies also $phi$, whatever the values of the domain $M$ of $mathcal M$ are assigned to the new constants $c_i$.
– Mauro ALLEGRANZA
Nov 28 '18 at 12:38
add a comment |
1 Answer
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The trick is that you didn't add any axioms about the constants in $C$.
If $T$ is inconsistent, it proves anything, so might as well assume that $T$ is consistent. Take any model of $T$, and any $m_1,ldots,m_n$ in that model, now interpret the constants $c_i$ as $m_i$ and any other constant symbol as $m_1$.
Since $T$ proved that $phi(c_1,ldots,c_n)$, it means that in $M$ (as an $L'$-structure), $phi(m_1,ldots,m_n)$ holds. But this is true for any such $n$-tuple in $M$. In particular, $Mmodelsforall x_1ldotsforall x_nphi(x_1,ldots,x_n)$. And now this is true for all $M$.
answered Nov 28 '18 at 12:36
Asaf Karagila♦
302k32425756
add a comment |
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The trick is that you didn't add any axioms about the constants in $C$.
If $T$ is inconsistent, it proves anything, so might as well assume that $T$ is consistent. Take any model of $T$, and any $m_1,ldots,m_n$ in that model, now interpret the constants $c_i$ as $m_i$ and any other constant symbol as $m_1$.
Since $T$ proved that $phi(c_1,ldots,c_n)$, it means that in $M$ (as an $L'$-structure), $phi(m_1,ldots,m_n)$ holds. But this is true for any such $n$-tuple in $M$. In particular, $Mmodelsforall x_1ldotsforall x_nphi(x_1,ldots,x_n)$. And now this is true for all $M$.
answered Nov 28 '18 at 12:36
Asaf Karagila♦
302k32425756
add a comment |
The trick is that you didn't add any axioms about the constants in $C$.
If $T$ is inconsistent, it proves anything, so might as well assume that $T$ is consistent. Take any model of $T$, and any $m_1,ldots,m_n$ in that model, now interpret the constants $c_i$ as $m_i$ and any other constant symbol as $m_1$.
Since $T$ proved that $phi(c_1,ldots,c_n)$, it means that in $M$ (as an $L'$-structure), $phi(m_1,ldots,m_n)$ holds. But this is true for any such $n$-tuple in $M$. In particular, $Mmodelsforall x_1ldotsforall x_nphi(x_1,ldots,x_n)$. And now this is true for all $M$.
answered Nov 28 '18 at 12:36
Asaf Karagila♦
302k32425756
add a comment |
The trick is that you didn't add any axioms about the constants in $C$.
If $T$ is inconsistent, it proves anything, so might as well assume that $T$ is consistent. Take any model of $T$, and any $m_1,ldots,m_n$ in that model, now interpret the constants $c_i$ as $m_i$ and any other constant symbol as $m_1$.
Since $T$ proved that $phi(c_1,ldots,c_n)$, it means that in $M$ (as an $L'$-structure), $phi(m_1,ldots,m_n)$ holds. But this is true for any such $n$-tuple in $M$. In particular, $Mmodelsforall x_1ldotsforall x_nphi(x_1,ldots,x_n)$. And now this is true for all $M$.
answered Nov 28 '18 at 12:36
Asaf Karagila♦
302k32425756
The trick is that you didn't add any axioms about the constants in $C$.
If $T$ is inconsistent, it proves anything, so might as well assume that $T$ is consistent. Take any model of $T$, and any $m_1,ldots,m_n$ in that model, now interpret the constants $c_i$ as $m_i$ and any other constant symbol as $m_1$.
Since $T$ proved that $phi(c_1,ldots,c_n)$, it means that in $M$ (as an $L'$-structure), $phi(m_1,ldots,m_n)$ holds. But this is true for any such $n$-tuple in $M$. In particular, $Mmodelsforall x_1ldotsforall x_nphi(x_1,ldots,x_n)$. And now this is true for all $M$.
answered Nov 28 '18 at 12:36
Asaf Karagila♦
302k32425756
answered Nov 28 '18 at 12:36
Asaf Karagila♦
302k32425756
answered Nov 28 '18 at 12:36
Asaf Karagila♦
302k32425756
answered Nov 28 '18 at 12:36
Asaf Karagila♦
302k32425756
302k32425756
add a comment |
add a comment |
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A theory is a set of sentences; thus, $T vDash phi$ means that sentence $phi$ is a logical cons of $T$. If $T$ ia an $L$-theory, a model $mathcal M$ for $T$ does not specifiy how to interpret the new constants $c_i$. Thus, $TvDash phi$ means that every model $mathcal M$ of $T$ satisfies also $phi$, whatever the values of the domain $M$ of $mathcal M$ are assigned to the new constants $c_i$.
– Mauro ALLEGRANZA
Nov 28 '18 at 12:38