Why is it if every model of $Sigma$ has an L'-expansion then $Sigma'$ is conservative over $Sigma$?
$begingroup$
Consider $Sigma subseteq Sigma'$ and $L subseteq L'$. Then the theorem is:
If every model of $Sigma$ has an L'-expansion, then $Sigma'$ is conservative over $Sigma$.
Recall conservative means: $forall sigma$ that are L-sentences,
$$ Sigma ' vdash_{L'} sigma iff Sigma vdash_L sigma $$
Recall that an L'-expansion means we extend the language only and not the set of the structures so: $mathcal A = (A,L)$ and $ mathcal A' = (A,L')$ is true and $L subseteq L'$.
It's clear that $Leftarrow$ is easy, since to form a proof we just copt the same proof from $Sigma$ and ignore any additional assumptions in our axioms $Sigma'$. Also, $L'$ is irrelevant since $sigma $ is an L-sentence.
I have a hint that we should do this proof by completeness. So this is what I tried:
WTS, $forall sigma$ L-sentences:
$$ Sigma ' vdash_{L'} sigma Rightarrow Sigma vdash_L sigma $$
So assume
1) $Sigma ' vdash_{L'} sigma$ is true.
By completeness we have every model $mathcal A'$ of $Sigma'$ models $sigma$ i.e. $mathcal A' models sigma$.
Ok but consider some model $mathcal A models Sigma$. Then by assumption there is an $L'$-extension $mathcal A' models Sigma' implies mathcal A' models sigma$.
Here is where I get stuck and can't connect to back to $mathcal A$. Why would $mathcal A models sigma$?
To me it seems I've reduced the problem to showing something like:
$$ mathcal A' models sigma implies mathcal A models sigma $$
which I would assume is always trivially true if $mathcal A'$ only contains additional symbols and the sets didn't change and if $sigma$ is an L-sentence. Since, $sigma$ only contain original symbols and the interpretations of those symbols didn't change so $ mathcal A' models sigma implies mathcal A models sigma $ must hold.
context:
https://faculty.math.illinois.edu/~vddries/main.pdf
logic first-order-logic model-theory
asked Dec 18 '18 at 3:11
PinocchioPinocchio
1,90521854
$endgroup$
add a comment |
$begingroup$
Consider $Sigma subseteq Sigma'$ and $L subseteq L'$. Then the theorem is:
If every model of $Sigma$ has an L'-expansion, then $Sigma'$ is conservative over $Sigma$.
Recall conservative means: $forall sigma$ that are L-sentences,
$$ Sigma ' vdash_{L'} sigma iff Sigma vdash_L sigma $$
Recall that an L'-expansion means we extend the language only and not the set of the structures so: $mathcal A = (A,L)$ and $ mathcal A' = (A,L')$ is true and $L subseteq L'$.
It's clear that $Leftarrow$ is easy, since to form a proof we just copt the same proof from $Sigma$ and ignore any additional assumptions in our axioms $Sigma'$. Also, $L'$ is irrelevant since $sigma $ is an L-sentence.
I have a hint that we should do this proof by completeness. So this is what I tried:
WTS, $forall sigma$ L-sentences:
$$ Sigma ' vdash_{L'} sigma Rightarrow Sigma vdash_L sigma $$
So assume
1) $Sigma ' vdash_{L'} sigma$ is true.
By completeness we have every model $mathcal A'$ of $Sigma'$ models $sigma$ i.e. $mathcal A' models sigma$.
Ok but consider some model $mathcal A models Sigma$. Then by assumption there is an $L'$-extension $mathcal A' models Sigma' implies mathcal A' models sigma$.
Here is where I get stuck and can't connect to back to $mathcal A$. Why would $mathcal A models sigma$?
To me it seems I've reduced the problem to showing something like:
$$ mathcal A' models sigma implies mathcal A models sigma $$
which I would assume is always trivially true if $mathcal A'$ only contains additional symbols and the sets didn't change and if $sigma$ is an L-sentence. Since, $sigma$ only contain original symbols and the interpretations of those symbols didn't change so $ mathcal A' models sigma implies mathcal A models sigma $ must hold.
context:
https://faculty.math.illinois.edu/~vddries/main.pdf
logic first-order-logic model-theory
asked Dec 18 '18 at 3:11
PinocchioPinocchio
1,90521854
$endgroup$
2
$begingroup$
You mean if every model of $Sigma$ has an $L'$-expansion to a model of $Sigma'$, then $Sigma'$ is conservative over $Sigma.$
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:09
2
$begingroup$
And yes, your reasoning in the last paragraph is correct: Since the domains are the same, the reduct of $mathcal A'$ to $L$ is $mathcal A,$ so they agree on $L$-sentences. I can't tell if you have a question beyond that.
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:26
add a comment |
$begingroup$
Consider $Sigma subseteq Sigma'$ and $L subseteq L'$. Then the theorem is:
If every model of $Sigma$ has an L'-expansion, then $Sigma'$ is conservative over $Sigma$.
Recall conservative means: $forall sigma$ that are L-sentences,
$$ Sigma ' vdash_{L'} sigma iff Sigma vdash_L sigma $$
Recall that an L'-expansion means we extend the language only and not the set of the structures so: $mathcal A = (A,L)$ and $ mathcal A' = (A,L')$ is true and $L subseteq L'$.
It's clear that $Leftarrow$ is easy, since to form a proof we just copt the same proof from $Sigma$ and ignore any additional assumptions in our axioms $Sigma'$. Also, $L'$ is irrelevant since $sigma $ is an L-sentence.
I have a hint that we should do this proof by completeness. So this is what I tried:
WTS, $forall sigma$ L-sentences:
$$ Sigma ' vdash_{L'} sigma Rightarrow Sigma vdash_L sigma $$
So assume
1) $Sigma ' vdash_{L'} sigma$ is true.
By completeness we have every model $mathcal A'$ of $Sigma'$ models $sigma$ i.e. $mathcal A' models sigma$.
Ok but consider some model $mathcal A models Sigma$. Then by assumption there is an $L'$-extension $mathcal A' models Sigma' implies mathcal A' models sigma$.
Here is where I get stuck and can't connect to back to $mathcal A$. Why would $mathcal A models sigma$?
To me it seems I've reduced the problem to showing something like:
$$ mathcal A' models sigma implies mathcal A models sigma $$
which I would assume is always trivially true if $mathcal A'$ only contains additional symbols and the sets didn't change and if $sigma$ is an L-sentence. Since, $sigma$ only contain original symbols and the interpretations of those symbols didn't change so $ mathcal A' models sigma implies mathcal A models sigma $ must hold.
context:
https://faculty.math.illinois.edu/~vddries/main.pdf
logic first-order-logic model-theory
asked Dec 18 '18 at 3:11
PinocchioPinocchio
1,90521854
$endgroup$
Consider $Sigma subseteq Sigma'$ and $L subseteq L'$. Then the theorem is:
If every model of $Sigma$ has an L'-expansion, then $Sigma'$ is conservative over $Sigma$.
Recall conservative means: $forall sigma$ that are L-sentences,
$$ Sigma ' vdash_{L'} sigma iff Sigma vdash_L sigma $$
Recall that an L'-expansion means we extend the language only and not the set of the structures so: $mathcal A = (A,L)$ and $ mathcal A' = (A,L')$ is true and $L subseteq L'$.
It's clear that $Leftarrow$ is easy, since to form a proof we just copt the same proof from $Sigma$ and ignore any additional assumptions in our axioms $Sigma'$. Also, $L'$ is irrelevant since $sigma $ is an L-sentence.
I have a hint that we should do this proof by completeness. So this is what I tried:
WTS, $forall sigma$ L-sentences:
$$ Sigma ' vdash_{L'} sigma Rightarrow Sigma vdash_L sigma $$
So assume
1) $Sigma ' vdash_{L'} sigma$ is true.
By completeness we have every model $mathcal A'$ of $Sigma'$ models $sigma$ i.e. $mathcal A' models sigma$.
Ok but consider some model $mathcal A models Sigma$. Then by assumption there is an $L'$-extension $mathcal A' models Sigma' implies mathcal A' models sigma$.
Here is where I get stuck and can't connect to back to $mathcal A$. Why would $mathcal A models sigma$?
To me it seems I've reduced the problem to showing something like:
$$ mathcal A' models sigma implies mathcal A models sigma $$
which I would assume is always trivially true if $mathcal A'$ only contains additional symbols and the sets didn't change and if $sigma$ is an L-sentence. Since, $sigma$ only contain original symbols and the interpretations of those symbols didn't change so $ mathcal A' models sigma implies mathcal A models sigma $ must hold.
context:
https://faculty.math.illinois.edu/~vddries/main.pdf
logic first-order-logic model-theory
logic first-order-logic model-theory
asked Dec 18 '18 at 3:11
PinocchioPinocchio
1,90521854
asked Dec 18 '18 at 3:11
PinocchioPinocchio
1,90521854
asked Dec 18 '18 at 3:11
PinocchioPinocchio
1,90521854
asked Dec 18 '18 at 3:11
PinocchioPinocchio
1,90521854
asked Dec 18 '18 at 3:11
PinocchioPinocchio
1,90521854
1,90521854
2
$begingroup$
You mean if every model of $Sigma$ has an $L'$-expansion to a model of $Sigma'$, then $Sigma'$ is conservative over $Sigma.$
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:09
2
$begingroup$
And yes, your reasoning in the last paragraph is correct: Since the domains are the same, the reduct of $mathcal A'$ to $L$ is $mathcal A,$ so they agree on $L$-sentences. I can't tell if you have a question beyond that.
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:26
add a comment |
2
$begingroup$
You mean if every model of $Sigma$ has an $L'$-expansion to a model of $Sigma'$, then $Sigma'$ is conservative over $Sigma.$
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:09
2
$begingroup$
And yes, your reasoning in the last paragraph is correct: Since the domains are the same, the reduct of $mathcal A'$ to $L$ is $mathcal A,$ so they agree on $L$-sentences. I can't tell if you have a question beyond that.
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:26
2
2
$begingroup$
You mean if every model of $Sigma$ has an $L'$-expansion to a model of $Sigma'$, then $Sigma'$ is conservative over $Sigma.$
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:09
$begingroup$
You mean if every model of $Sigma$ has an $L'$-expansion to a model of $Sigma'$, then $Sigma'$ is conservative over $Sigma.$
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:09
2
2
$begingroup$
And yes, your reasoning in the last paragraph is correct: Since the domains are the same, the reduct of $mathcal A'$ to $L$ is $mathcal A,$ so they agree on $L$-sentences. I can't tell if you have a question beyond that.
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:26
$begingroup$
And yes, your reasoning in the last paragraph is correct: Since the domains are the same, the reduct of $mathcal A'$ to $L$ is $mathcal A,$ so they agree on $L$-sentences. I can't tell if you have a question beyond that.
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:26
add a comment |
0
active
oldest
votes
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%2f3044739%2fwhy-is-it-if-every-model-of-sigma-has-an-l-expansion-then-sigma-is-conse%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3044739%2fwhy-is-it-if-every-model-of-sigma-has-an-l-expansion-then-sigma-is-conse%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
2
$begingroup$
You mean if every model of $Sigma$ has an $L'$-expansion to a model of $Sigma'$, then $Sigma'$ is conservative over $Sigma.$
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:09
2
$begingroup$
And yes, your reasoning in the last paragraph is correct: Since the domains are the same, the reduct of $mathcal A'$ to $L$ is $mathcal A,$ so they agree on $L$-sentences. I can't tell if you have a question beyond that.
$endgroup$
– spaceisdarkgreen
Dec 18 '18 at 5:26