$exists^infty$-elimination and model companion.
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In the book Ziegler, Tent: A course in model theory it states
Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.
I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise
Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
$T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.
I define the theory
$(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$
To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?
model-theory
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In the book Ziegler, Tent: A course in model theory it states
Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.
I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise
Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
$T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.
I define the theory
$(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$
To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?
model-theory
add a comment |
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In the book Ziegler, Tent: A course in model theory it states
Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.
I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise
Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
$T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.
I define the theory
$(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$
To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?
model-theory
In the book Ziegler, Tent: A course in model theory it states
Exercise 5.5.7. Let $T_1$ and $T_2$ be two model complete theories in disjoint
languages $L_1$ and $L_2$. Assume that both theories eliminate $exists^infty$ . Then $T_1cup T_2$ has a model companion.
I want to prove this statement. Sine $T_1cup T_2$ also eliminates $exists^infty$ (i.e. in all models realisation sets have either a finite upper bound or are infinite) my strategy was to use the previous exercise
Exercise 5.5.6. Assume that T eliminates the quantifier $exists^infty$. Then for every formula $varphi(x_1 , . . . , x_n , overline{y})$ there is a formula $theta(overline{y})$ such that in all models $mathfrak{M}$ of
$T$ a tuple $overline{b}$ satisfies $theta(overline{y})$ if and only if $mathfrak{M}$ has an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$.
I define the theory
$(T_1cup T_2)^*:={theta;|;$there exists a $L_1cup L_2$-formula $varphi(x_1,..x_n)$, that is satisfiable in all models $mathfrak{M}vDash T_1cup T_2$ such that $mathfrak{M}vDashthetaLeftrightarrow$ there exists an elementary extension $mathfrak{M}'$ with
elements $a_1 , . . . , a_nin M'setminus M$ such that $mathfrak{M}vDashvarphi(a_1,...,a_n)$$}$
To me it seemed to be right choice for the model companion of $T_1cup T_2$ until I tried to prove that $(T_1cup T_2)^*$ is model complete (without success). Does this choice make sense?
model-theory
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