Conjunction closure of positive primitive formulas
$begingroup$
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...
logic first-order-logic model-theory
$endgroup$
add a comment |
$begingroup$
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...
logic first-order-logic model-theory
$endgroup$
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
add a comment |
$begingroup$
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...
logic first-order-logic model-theory
$endgroup$
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
logic first-order-logic model-theory
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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:
- If the pp-formulas have any quantified variables in common, rename these variables in one of the formulas. This avoids "mixing up existentials".
- Expand all equations to include all variables, by attaching zero coefficients to the new variables.
- Take the conjunction of all of the equations, and put the existential quantifiers in front.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045314%2fconjunction-closure-of-positive-primitive-formulas%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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:
- If the pp-formulas have any quantified variables in common, rename these variables in one of the formulas. This avoids "mixing up existentials".
- Expand all equations to include all variables, by attaching zero coefficients to the new variables.
- Take the conjunction of all of the equations, and put the existential quantifiers in front.
$endgroup$
add a comment |
$begingroup$
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:
- If the pp-formulas have any quantified variables in common, rename these variables in one of the formulas. This avoids "mixing up existentials".
- Expand all equations to include all variables, by attaching zero coefficients to the new variables.
- Take the conjunction of all of the equations, and put the existential quantifiers in front.
$endgroup$
add a comment |
$begingroup$
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:
- If the pp-formulas have any quantified variables in common, rename these variables in one of the formulas. This avoids "mixing up existentials".
- Expand all equations to include all variables, by attaching zero coefficients to the new variables.
- Take the conjunction of all of the equations, and put the existential quantifiers in front.
$endgroup$
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:
- If the pp-formulas have any quantified variables in common, rename these variables in one of the formulas. This avoids "mixing up existentials".
- Expand all equations to include all variables, by attaching zero coefficients to the new variables.
- Take the conjunction of all of the equations, and put the existential quantifiers in front.
answered Dec 18 '18 at 16:54
Alex KruckmanAlex Kruckman
27.6k32658
27.6k32658
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045314%2fconjunction-closure-of-positive-primitive-formulas%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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