Conjunction closure of positive primitive formulas












0












$begingroup$


Let $mathfrak M = (M, 0, +, - , r)_{rin R}$ be a structure which realizes a module, i.e., $(M, 0, +, -)$ is an abelian group together with operations $r : M to M$ with $R$ a ring with $1$. Let $L_{Mod}(R) = L_{AbG} cup {r : r in R }$ with $L_{AbG} = {0,+,-}$ be the language of $R$-modules.



Definition: An equation is an $L_{Mod}(R)$-formula $gamma(overline x)$ of the form
$$
r_1 x_1 + r_2 x_2 + ldots + r_m x_m = 0.
$$

A positive primitive formula ($pp$-formula) is of the form
$$
exists overline y (gamma_1 land ldots land gamma_n)
$$

where $gamma_i(overline x, overline y)$ are equations. These definitions are taken form K.Tent/M.Ziegler, A Course in Model Theory. In these notes it is mentioned that we can combine $pp$-formulas with intersection. If $varphi(overline x)$ and $psi(overline x)$ are pp-formuals then
$$
(varphicap psi)(overline x) = varphi(overline x) land psi(overline x).
$$

But why would this give a $pp$-formula? For example if
$$
varphi(x) = exists y ( r_1 x + r_2 y = 0 ), quad
psi(x) = exists y ( s_1 x + s_2 y = 0 )
$$

then
$$
exists y ( r_1 x + r_2 y = 0 ) land
exists y ( s_1 x + s_2 y = 0 )
$$

is not a pp-formula as the existential quantifier does not appears at the beginning? Maybe we could write
$$
exists y exists z ( r_1 x + r_2 y = 0 land s_1 x + s_2 z = 0 )
$$

but this is not what is suggested in the text...










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$endgroup$








  • 1




    $begingroup$
    Do you know what is the prenex normal form of a formula? en.wikipedia.org/wiki/Prenex_normal_form
    $endgroup$
    – Gödel
    Dec 18 '18 at 16:46






  • 1




    $begingroup$
    I fixed your title from "conjugation closure" to "conjunction closure".
    $endgroup$
    – Alex Kruckman
    Dec 18 '18 at 17:07






  • 1




    $begingroup$
    @Gödel Thanks, I see it is basically entails the formula Alex Kruckman suggested in his answer.
    $endgroup$
    – StefanH
    Dec 18 '18 at 17:14










  • $begingroup$
    @AlexKruckman Yes, that was a typo!
    $endgroup$
    – StefanH
    Dec 18 '18 at 17:14
















0












$begingroup$


Let $mathfrak M = (M, 0, +, - , r)_{rin R}$ be a structure which realizes a module, i.e., $(M, 0, +, -)$ is an abelian group together with operations $r : M to M$ with $R$ a ring with $1$. Let $L_{Mod}(R) = L_{AbG} cup {r : r in R }$ with $L_{AbG} = {0,+,-}$ be the language of $R$-modules.



Definition: An equation is an $L_{Mod}(R)$-formula $gamma(overline x)$ of the form
$$
r_1 x_1 + r_2 x_2 + ldots + r_m x_m = 0.
$$

A positive primitive formula ($pp$-formula) is of the form
$$
exists overline y (gamma_1 land ldots land gamma_n)
$$

where $gamma_i(overline x, overline y)$ are equations. These definitions are taken form K.Tent/M.Ziegler, A Course in Model Theory. In these notes it is mentioned that we can combine $pp$-formulas with intersection. If $varphi(overline x)$ and $psi(overline x)$ are pp-formuals then
$$
(varphicap psi)(overline x) = varphi(overline x) land psi(overline x).
$$

But why would this give a $pp$-formula? For example if
$$
varphi(x) = exists y ( r_1 x + r_2 y = 0 ), quad
psi(x) = exists y ( s_1 x + s_2 y = 0 )
$$

then
$$
exists y ( r_1 x + r_2 y = 0 ) land
exists y ( s_1 x + s_2 y = 0 )
$$

is not a pp-formula as the existential quantifier does not appears at the beginning? Maybe we could write
$$
exists y exists z ( r_1 x + r_2 y = 0 land s_1 x + s_2 z = 0 )
$$

but this is not what is suggested in the text...










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Do you know what is the prenex normal form of a formula? en.wikipedia.org/wiki/Prenex_normal_form
    $endgroup$
    – Gödel
    Dec 18 '18 at 16:46






  • 1




    $begingroup$
    I fixed your title from "conjugation closure" to "conjunction closure".
    $endgroup$
    – Alex Kruckman
    Dec 18 '18 at 17:07






  • 1




    $begingroup$
    @Gödel Thanks, I see it is basically entails the formula Alex Kruckman suggested in his answer.
    $endgroup$
    – StefanH
    Dec 18 '18 at 17:14










  • $begingroup$
    @AlexKruckman Yes, that was a typo!
    $endgroup$
    – StefanH
    Dec 18 '18 at 17:14














0












0








0





$begingroup$


Let $mathfrak M = (M, 0, +, - , r)_{rin R}$ be a structure which realizes a module, i.e., $(M, 0, +, -)$ is an abelian group together with operations $r : M to M$ with $R$ a ring with $1$. Let $L_{Mod}(R) = L_{AbG} cup {r : r in R }$ with $L_{AbG} = {0,+,-}$ be the language of $R$-modules.



Definition: An equation is an $L_{Mod}(R)$-formula $gamma(overline x)$ of the form
$$
r_1 x_1 + r_2 x_2 + ldots + r_m x_m = 0.
$$

A positive primitive formula ($pp$-formula) is of the form
$$
exists overline y (gamma_1 land ldots land gamma_n)
$$

where $gamma_i(overline x, overline y)$ are equations. These definitions are taken form K.Tent/M.Ziegler, A Course in Model Theory. In these notes it is mentioned that we can combine $pp$-formulas with intersection. If $varphi(overline x)$ and $psi(overline x)$ are pp-formuals then
$$
(varphicap psi)(overline x) = varphi(overline x) land psi(overline x).
$$

But why would this give a $pp$-formula? For example if
$$
varphi(x) = exists y ( r_1 x + r_2 y = 0 ), quad
psi(x) = exists y ( s_1 x + s_2 y = 0 )
$$

then
$$
exists y ( r_1 x + r_2 y = 0 ) land
exists y ( s_1 x + s_2 y = 0 )
$$

is not a pp-formula as the existential quantifier does not appears at the beginning? Maybe we could write
$$
exists y exists z ( r_1 x + r_2 y = 0 land s_1 x + s_2 z = 0 )
$$

but this is not what is suggested in the text...










share|cite|improve this question











$endgroup$




Let $mathfrak M = (M, 0, +, - , r)_{rin R}$ be a structure which realizes a module, i.e., $(M, 0, +, -)$ is an abelian group together with operations $r : M to M$ with $R$ a ring with $1$. Let $L_{Mod}(R) = L_{AbG} cup {r : r in R }$ with $L_{AbG} = {0,+,-}$ be the language of $R$-modules.



Definition: An equation is an $L_{Mod}(R)$-formula $gamma(overline x)$ of the form
$$
r_1 x_1 + r_2 x_2 + ldots + r_m x_m = 0.
$$

A positive primitive formula ($pp$-formula) is of the form
$$
exists overline y (gamma_1 land ldots land gamma_n)
$$

where $gamma_i(overline x, overline y)$ are equations. These definitions are taken form K.Tent/M.Ziegler, A Course in Model Theory. In these notes it is mentioned that we can combine $pp$-formulas with intersection. If $varphi(overline x)$ and $psi(overline x)$ are pp-formuals then
$$
(varphicap psi)(overline x) = varphi(overline x) land psi(overline x).
$$

But why would this give a $pp$-formula? For example if
$$
varphi(x) = exists y ( r_1 x + r_2 y = 0 ), quad
psi(x) = exists y ( s_1 x + s_2 y = 0 )
$$

then
$$
exists y ( r_1 x + r_2 y = 0 ) land
exists y ( s_1 x + s_2 y = 0 )
$$

is not a pp-formula as the existential quantifier does not appears at the beginning? Maybe we could write
$$
exists y exists z ( r_1 x + r_2 y = 0 land s_1 x + s_2 z = 0 )
$$

but this is not what is suggested in the text...







logic first-order-logic model-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 17:06









Alex Kruckman

27.6k32658




27.6k32658










asked Dec 18 '18 at 15:47









StefanHStefanH

8,14652366




8,14652366








  • 1




    $begingroup$
    Do you know what is the prenex normal form of a formula? en.wikipedia.org/wiki/Prenex_normal_form
    $endgroup$
    – Gödel
    Dec 18 '18 at 16:46






  • 1




    $begingroup$
    I fixed your title from "conjugation closure" to "conjunction closure".
    $endgroup$
    – Alex Kruckman
    Dec 18 '18 at 17:07






  • 1




    $begingroup$
    @Gödel Thanks, I see it is basically entails the formula Alex Kruckman suggested in his answer.
    $endgroup$
    – StefanH
    Dec 18 '18 at 17:14










  • $begingroup$
    @AlexKruckman Yes, that was a typo!
    $endgroup$
    – StefanH
    Dec 18 '18 at 17:14














  • 1




    $begingroup$
    Do you know what is the prenex normal form of a formula? en.wikipedia.org/wiki/Prenex_normal_form
    $endgroup$
    – Gödel
    Dec 18 '18 at 16:46






  • 1




    $begingroup$
    I fixed your title from "conjugation closure" to "conjunction closure".
    $endgroup$
    – Alex Kruckman
    Dec 18 '18 at 17:07






  • 1




    $begingroup$
    @Gödel Thanks, I see it is basically entails the formula Alex Kruckman suggested in his answer.
    $endgroup$
    – StefanH
    Dec 18 '18 at 17:14










  • $begingroup$
    @AlexKruckman Yes, that was a typo!
    $endgroup$
    – StefanH
    Dec 18 '18 at 17:14








1




1




$begingroup$
Do you know what is the prenex normal form of a formula? en.wikipedia.org/wiki/Prenex_normal_form
$endgroup$
– Gödel
Dec 18 '18 at 16:46




$begingroup$
Do you know what is the prenex normal form of a formula? en.wikipedia.org/wiki/Prenex_normal_form
$endgroup$
– Gödel
Dec 18 '18 at 16:46




1




1




$begingroup$
I fixed your title from "conjugation closure" to "conjunction closure".
$endgroup$
– Alex Kruckman
Dec 18 '18 at 17:07




$begingroup$
I fixed your title from "conjugation closure" to "conjunction closure".
$endgroup$
– Alex Kruckman
Dec 18 '18 at 17:07




1




1




$begingroup$
@Gödel Thanks, I see it is basically entails the formula Alex Kruckman suggested in his answer.
$endgroup$
– StefanH
Dec 18 '18 at 17:14




$begingroup$
@Gödel Thanks, I see it is basically entails the formula Alex Kruckman suggested in his answer.
$endgroup$
– StefanH
Dec 18 '18 at 17:14












$begingroup$
@AlexKruckman Yes, that was a typo!
$endgroup$
– StefanH
Dec 18 '18 at 17:14




$begingroup$
@AlexKruckman Yes, that was a typo!
$endgroup$
– StefanH
Dec 18 '18 at 17:14










1 Answer
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$begingroup$

The sentence you wrote at the end is correct. Although formally, you should maybe write $$exists y, exists z, (r_1x+r_2y+0z = 0 land s_1x+0y+s_2z = 0)$$ because the definition says a pp-formula should have the form $exists z, (gamma_1(x,z)land dotsland gamma_n(x,z))$, where the $gamma_i$ are equations in the variables $xz$, i.e. they include all of the free and quantified variables.



You say that this isn't suggested in the text, but I think it is - sometimes you have to read between the lines a little!



First, you're expected to know the basic logic fact that the conjunction of two existential formulas is equivalent to an existential formula, but you have to rename any conflicting variables: $$(exists y, varphi(x,y))land (exists y, psi(x,y)) Leftrightarrow exists y, exists z, (varphi(x,y)land psi(x,z)).$$



Now the authors of the linked document write: "We can see any equation $gamma(x,y)$ in some variables as an equation $tilde{gamma}(x,y,z)$ relating more variables, simply attaching zero coefficients to the new variables. Hence we can combine two pp-formulas without mixing up existentials."



So they're suggesting the following procedure:




  1. If the pp-formulas have any quantified variables in common, rename these variables in one of the formulas. This avoids "mixing up existentials".

  2. Expand all equations to include all variables, by attaching zero coefficients to the new variables.

  3. Take the conjunction of all of the equations, and put the existential quantifiers in front.






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    $begingroup$

    The sentence you wrote at the end is correct. Although formally, you should maybe write $$exists y, exists z, (r_1x+r_2y+0z = 0 land s_1x+0y+s_2z = 0)$$ because the definition says a pp-formula should have the form $exists z, (gamma_1(x,z)land dotsland gamma_n(x,z))$, where the $gamma_i$ are equations in the variables $xz$, i.e. they include all of the free and quantified variables.



    You say that this isn't suggested in the text, but I think it is - sometimes you have to read between the lines a little!



    First, you're expected to know the basic logic fact that the conjunction of two existential formulas is equivalent to an existential formula, but you have to rename any conflicting variables: $$(exists y, varphi(x,y))land (exists y, psi(x,y)) Leftrightarrow exists y, exists z, (varphi(x,y)land psi(x,z)).$$



    Now the authors of the linked document write: "We can see any equation $gamma(x,y)$ in some variables as an equation $tilde{gamma}(x,y,z)$ relating more variables, simply attaching zero coefficients to the new variables. Hence we can combine two pp-formulas without mixing up existentials."



    So they're suggesting the following procedure:




    1. If the pp-formulas have any quantified variables in common, rename these variables in one of the formulas. This avoids "mixing up existentials".

    2. Expand all equations to include all variables, by attaching zero coefficients to the new variables.

    3. Take the conjunction of all of the equations, and put the existential quantifiers in front.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The sentence you wrote at the end is correct. Although formally, you should maybe write $$exists y, exists z, (r_1x+r_2y+0z = 0 land s_1x+0y+s_2z = 0)$$ because the definition says a pp-formula should have the form $exists z, (gamma_1(x,z)land dotsland gamma_n(x,z))$, where the $gamma_i$ are equations in the variables $xz$, i.e. they include all of the free and quantified variables.



      You say that this isn't suggested in the text, but I think it is - sometimes you have to read between the lines a little!



      First, you're expected to know the basic logic fact that the conjunction of two existential formulas is equivalent to an existential formula, but you have to rename any conflicting variables: $$(exists y, varphi(x,y))land (exists y, psi(x,y)) Leftrightarrow exists y, exists z, (varphi(x,y)land psi(x,z)).$$



      Now the authors of the linked document write: "We can see any equation $gamma(x,y)$ in some variables as an equation $tilde{gamma}(x,y,z)$ relating more variables, simply attaching zero coefficients to the new variables. Hence we can combine two pp-formulas without mixing up existentials."



      So they're suggesting the following procedure:




      1. If the pp-formulas have any quantified variables in common, rename these variables in one of the formulas. This avoids "mixing up existentials".

      2. Expand all equations to include all variables, by attaching zero coefficients to the new variables.

      3. Take the conjunction of all of the equations, and put the existential quantifiers in front.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The sentence you wrote at the end is correct. Although formally, you should maybe write $$exists y, exists z, (r_1x+r_2y+0z = 0 land s_1x+0y+s_2z = 0)$$ because the definition says a pp-formula should have the form $exists z, (gamma_1(x,z)land dotsland gamma_n(x,z))$, where the $gamma_i$ are equations in the variables $xz$, i.e. they include all of the free and quantified variables.



        You say that this isn't suggested in the text, but I think it is - sometimes you have to read between the lines a little!



        First, you're expected to know the basic logic fact that the conjunction of two existential formulas is equivalent to an existential formula, but you have to rename any conflicting variables: $$(exists y, varphi(x,y))land (exists y, psi(x,y)) Leftrightarrow exists y, exists z, (varphi(x,y)land psi(x,z)).$$



        Now the authors of the linked document write: "We can see any equation $gamma(x,y)$ in some variables as an equation $tilde{gamma}(x,y,z)$ relating more variables, simply attaching zero coefficients to the new variables. Hence we can combine two pp-formulas without mixing up existentials."



        So they're suggesting the following procedure:




        1. If the pp-formulas have any quantified variables in common, rename these variables in one of the formulas. This avoids "mixing up existentials".

        2. Expand all equations to include all variables, by attaching zero coefficients to the new variables.

        3. Take the conjunction of all of the equations, and put the existential quantifiers in front.






        share|cite|improve this answer









        $endgroup$



        The sentence you wrote at the end is correct. Although formally, you should maybe write $$exists y, exists z, (r_1x+r_2y+0z = 0 land s_1x+0y+s_2z = 0)$$ because the definition says a pp-formula should have the form $exists z, (gamma_1(x,z)land dotsland gamma_n(x,z))$, where the $gamma_i$ are equations in the variables $xz$, i.e. they include all of the free and quantified variables.



        You say that this isn't suggested in the text, but I think it is - sometimes you have to read between the lines a little!



        First, you're expected to know the basic logic fact that the conjunction of two existential formulas is equivalent to an existential formula, but you have to rename any conflicting variables: $$(exists y, varphi(x,y))land (exists y, psi(x,y)) Leftrightarrow exists y, exists z, (varphi(x,y)land psi(x,z)).$$



        Now the authors of the linked document write: "We can see any equation $gamma(x,y)$ in some variables as an equation $tilde{gamma}(x,y,z)$ relating more variables, simply attaching zero coefficients to the new variables. Hence we can combine two pp-formulas without mixing up existentials."



        So they're suggesting the following procedure:




        1. If the pp-formulas have any quantified variables in common, rename these variables in one of the formulas. This avoids "mixing up existentials".

        2. Expand all equations to include all variables, by attaching zero coefficients to the new variables.

        3. Take the conjunction of all of the equations, and put the existential quantifiers in front.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 16:54









        Alex KruckmanAlex Kruckman

        27.6k32658




        27.6k32658






























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