Isar proof of conjunction2019 Community Moderator ElectionWhat is a good way to define a finite multiplication table in Isar?When would you use `presume` in an Isar proof?Need proof for meta-logic conjunction elimination ruleTrying to understand fix/assume/show “Failure to refine goal”; Cmd to show proof info for schematic varsHow to submit an argument to keyword proof?Taming meta implication in Isar proofsExpressing a simple declarative proof about exponents in IsarProving properties of generated listsKnowing when an Isar-style proof is actually valid in IsabelleFunctor-like construction in Isabelle/Isar

How does a sound wave propagate?

Insult for someone who "doesn't know anything"

Is divide-by-zero a security vulnerability?

Why would /etc/passwd be used every time someone executes `ls -l` command?

Where is the License file location for Identity Server in Sitecore 9.1?

Who has more? Ireland or Iceland?

What is better: yes / no radio, or simple checkbox?

How to write a chaotic neutral protagonist and prevent my readers from thinking they are evil?

3.5% Interest Student Loan or use all of my savings on Tuition?

The (Easy) Road to Code

How to recover against Snake as a heavyweight character?

What do you call someone who likes to pick fights?

Use Mercury as quenching liquid for swords?

Is it a Cyclops number? "Nobody" knows!

Ultrafilters as a double dual

Short story about cities being connected by a conveyor belt

What can I do if someone tampers with my SSH public key?

How does learning spells work when leveling a multiclass character?

Generating a list with duplicate entries

Exempt portion of equation line from aligning?

What is the orbit and expected lifetime of Crew Dragon trunk?

Why aren't there more Gauls like Obelix?

A running toilet that stops itself

Should we avoid writing fiction about historical events without extensive research?



Isar proof of conjunction



2019 Community Moderator ElectionWhat is a good way to define a finite multiplication table in Isar?When would you use `presume` in an Isar proof?Need proof for meta-logic conjunction elimination ruleTrying to understand fix/assume/show “Failure to refine goal”; Cmd to show proof info for schematic varsHow to submit an argument to keyword proof?Taming meta implication in Isar proofsExpressing a simple declarative proof about exponents in IsarProving properties of generated listsKnowing when an Isar-style proof is actually valid in IsabelleFunctor-like construction in Isabelle/Isar










0















I'm trying to prove something using Isar; so far, I've gotten to a goal that looks like this:



(∀P Q. P ≠ Q ⟶ (∃!l. plmeets P l ∧ plmeets Q l)) ∧
(∀P l. ¬ plmeets P l ⟶ (∃!m. affine_plane_data.parallel plmeets l m ∧ plmeets P m)) ∧
(∃P Q. P ≠ Q ∧ (∃R. P ≠ R ∧ Q ≠ R ∧ ¬ affine_plane_data.collinear plmeets P Q R))


(here plmeets is a function I've defined, where plmeets P l is shorthand for "the point P lies on the line l" in an affine plane, but I don't think that's important to my question.)



This goal is a conjunction of three things. I've actually already proved lemmas that seem to me to be very close to each of these things. For instance, I have



lemma four_points_a1: "P ≠ Q ⟹ ∃! l . plmeets P l ∧ plmeets Q l"


which produces the output



theorem four_points_a1: ?P ≠ ?Q ⟹ ∃!l. plmeets ?P l ∧ plmeets ?Q l


which you can see is almost exactly the first of the three conjoined items. (I admit that my other lemmas aren't quite so exactly matched to the other two items, but I'll work on that).



I'd like to say "because of lemma four_points_a1, all we have left to prove is item2 ∧ item3" and I'm pretty sure there's a way to do this. But looking at the "Programming and Proving" book doesn't suggest anything to me. In Isabelle, rather than Isar, I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.



But I cannot see how to do this in Isar.










share|improve this question



















  • 1





    " I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal." It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.

    – xanonec
    2 days ago












  • Thanks. This appears to be the thing I needed to get me past this particular roadblock. I've posted it as a community-wiki answer so that I can "accept" it in a few days, leaving fewer unanswered questions lying around. (If you'd rather write it as an answer, I'll accept that one instead).

    – John
    2 days ago















0















I'm trying to prove something using Isar; so far, I've gotten to a goal that looks like this:



(∀P Q. P ≠ Q ⟶ (∃!l. plmeets P l ∧ plmeets Q l)) ∧
(∀P l. ¬ plmeets P l ⟶ (∃!m. affine_plane_data.parallel plmeets l m ∧ plmeets P m)) ∧
(∃P Q. P ≠ Q ∧ (∃R. P ≠ R ∧ Q ≠ R ∧ ¬ affine_plane_data.collinear plmeets P Q R))


(here plmeets is a function I've defined, where plmeets P l is shorthand for "the point P lies on the line l" in an affine plane, but I don't think that's important to my question.)



This goal is a conjunction of three things. I've actually already proved lemmas that seem to me to be very close to each of these things. For instance, I have



lemma four_points_a1: "P ≠ Q ⟹ ∃! l . plmeets P l ∧ plmeets Q l"


which produces the output



theorem four_points_a1: ?P ≠ ?Q ⟹ ∃!l. plmeets ?P l ∧ plmeets ?Q l


which you can see is almost exactly the first of the three conjoined items. (I admit that my other lemmas aren't quite so exactly matched to the other two items, but I'll work on that).



I'd like to say "because of lemma four_points_a1, all we have left to prove is item2 ∧ item3" and I'm pretty sure there's a way to do this. But looking at the "Programming and Proving" book doesn't suggest anything to me. In Isabelle, rather than Isar, I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.



But I cannot see how to do this in Isar.










share|improve this question



















  • 1





    " I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal." It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.

    – xanonec
    2 days ago












  • Thanks. This appears to be the thing I needed to get me past this particular roadblock. I've posted it as a community-wiki answer so that I can "accept" it in a few days, leaving fewer unanswered questions lying around. (If you'd rather write it as an answer, I'll accept that one instead).

    – John
    2 days ago













0












0








0








I'm trying to prove something using Isar; so far, I've gotten to a goal that looks like this:



(∀P Q. P ≠ Q ⟶ (∃!l. plmeets P l ∧ plmeets Q l)) ∧
(∀P l. ¬ plmeets P l ⟶ (∃!m. affine_plane_data.parallel plmeets l m ∧ plmeets P m)) ∧
(∃P Q. P ≠ Q ∧ (∃R. P ≠ R ∧ Q ≠ R ∧ ¬ affine_plane_data.collinear plmeets P Q R))


(here plmeets is a function I've defined, where plmeets P l is shorthand for "the point P lies on the line l" in an affine plane, but I don't think that's important to my question.)



This goal is a conjunction of three things. I've actually already proved lemmas that seem to me to be very close to each of these things. For instance, I have



lemma four_points_a1: "P ≠ Q ⟹ ∃! l . plmeets P l ∧ plmeets Q l"


which produces the output



theorem four_points_a1: ?P ≠ ?Q ⟹ ∃!l. plmeets ?P l ∧ plmeets ?Q l


which you can see is almost exactly the first of the three conjoined items. (I admit that my other lemmas aren't quite so exactly matched to the other two items, but I'll work on that).



I'd like to say "because of lemma four_points_a1, all we have left to prove is item2 ∧ item3" and I'm pretty sure there's a way to do this. But looking at the "Programming and Proving" book doesn't suggest anything to me. In Isabelle, rather than Isar, I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.



But I cannot see how to do this in Isar.










share|improve this question
















I'm trying to prove something using Isar; so far, I've gotten to a goal that looks like this:



(∀P Q. P ≠ Q ⟶ (∃!l. plmeets P l ∧ plmeets Q l)) ∧
(∀P l. ¬ plmeets P l ⟶ (∃!m. affine_plane_data.parallel plmeets l m ∧ plmeets P m)) ∧
(∃P Q. P ≠ Q ∧ (∃R. P ≠ R ∧ Q ≠ R ∧ ¬ affine_plane_data.collinear plmeets P Q R))


(here plmeets is a function I've defined, where plmeets P l is shorthand for "the point P lies on the line l" in an affine plane, but I don't think that's important to my question.)



This goal is a conjunction of three things. I've actually already proved lemmas that seem to me to be very close to each of these things. For instance, I have



lemma four_points_a1: "P ≠ Q ⟹ ∃! l . plmeets P l ∧ plmeets Q l"


which produces the output



theorem four_points_a1: ?P ≠ ?Q ⟹ ∃!l. plmeets ?P l ∧ plmeets ?Q l


which you can see is almost exactly the first of the three conjoined items. (I admit that my other lemmas aren't quite so exactly matched to the other two items, but I'll work on that).



I'd like to say "because of lemma four_points_a1, all we have left to prove is item2 ∧ item3" and I'm pretty sure there's a way to do this. But looking at the "Programming and Proving" book doesn't suggest anything to me. In Isabelle, rather than Isar, I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.



But I cannot see how to do this in Isar.







isabelle isar






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday







John

















asked 2 days ago









JohnJohn

246214




246214







  • 1





    " I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal." It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.

    – xanonec
    2 days ago












  • Thanks. This appears to be the thing I needed to get me past this particular roadblock. I've posted it as a community-wiki answer so that I can "accept" it in a few days, leaving fewer unanswered questions lying around. (If you'd rather write it as an answer, I'll accept that one instead).

    – John
    2 days ago












  • 1





    " I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal." It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.

    – xanonec
    2 days ago












  • Thanks. This appears to be the thing I needed to get me past this particular roadblock. I've posted it as a community-wiki answer so that I can "accept" it in a few days, leaving fewer unanswered questions lying around. (If you'd rather write it as an answer, I'll accept that one instead).

    – John
    2 days ago







1




1





" I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal." It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.

– xanonec
2 days ago






" I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal." It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.

– xanonec
2 days ago














Thanks. This appears to be the thing I needed to get me past this particular roadblock. I've posted it as a community-wiki answer so that I can "accept" it in a few days, leaving fewer unanswered questions lying around. (If you'd rather write it as an answer, I'll accept that one instead).

– John
2 days ago





Thanks. This appears to be the thing I needed to get me past this particular roadblock. I've posted it as a community-wiki answer so that I can "accept" it in a few days, leaving fewer unanswered questions lying around. (If you'd rather write it as an answer, I'll accept that one instead).

– John
2 days ago












1 Answer
1






active

oldest

votes


















0














According to @xanonec:




I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.




It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.




According to @John:
The syntax for this process that actually worked was this:



 proposition four_points_sufficient: "affine_plane plmeets"
unfolding affine_plane_def
apply (intro conjI)
subgoal using four_points_a1 by blast


It's not clear to me how "It is possible to do this [i.e., apply conjI twice] within an Isar proof," but perhaps I don't now need to know.






share|improve this answer




















  • 1





    "It is possible to do this [i.e., apply conjI twice] within an Isar proof,". It is possible to use a comma-separated list of methods (sequential composition, also see 6.4 in the reference manual) both in Isar and apply scripts, e.g lemma "True" proof fix A B :: "'a ⇒ bool" have "∀x y. A x ∧ B y ⟶ B y ∧ A x" by (rule allI, rule allI, simp) ....

    – xanonec
    yesterday






  • 1





    Also, there is nothing that could prevent one from using apply style script inside a proof block. To the best of my knowledge, it is not considered to be a very good practice, but it is not considered to be a particularly bad practice either (see "Do not switch from apply into proof in the middle of a proof" in https://proofcraft.org/blog/isabelle-style.html).

    – xanonec
    yesterday











Your Answer






StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
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
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55028088%2fisar-proof-of-conjunction%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









0














According to @xanonec:




I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.




It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.




According to @John:
The syntax for this process that actually worked was this:



 proposition four_points_sufficient: "affine_plane plmeets"
unfolding affine_plane_def
apply (intro conjI)
subgoal using four_points_a1 by blast


It's not clear to me how "It is possible to do this [i.e., apply conjI twice] within an Isar proof," but perhaps I don't now need to know.






share|improve this answer




















  • 1





    "It is possible to do this [i.e., apply conjI twice] within an Isar proof,". It is possible to use a comma-separated list of methods (sequential composition, also see 6.4 in the reference manual) both in Isar and apply scripts, e.g lemma "True" proof fix A B :: "'a ⇒ bool" have "∀x y. A x ∧ B y ⟶ B y ∧ A x" by (rule allI, rule allI, simp) ....

    – xanonec
    yesterday






  • 1





    Also, there is nothing that could prevent one from using apply style script inside a proof block. To the best of my knowledge, it is not considered to be a very good practice, but it is not considered to be a particularly bad practice either (see "Do not switch from apply into proof in the middle of a proof" in https://proofcraft.org/blog/isabelle-style.html).

    – xanonec
    yesterday
















0














According to @xanonec:




I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.




It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.




According to @John:
The syntax for this process that actually worked was this:



 proposition four_points_sufficient: "affine_plane plmeets"
unfolding affine_plane_def
apply (intro conjI)
subgoal using four_points_a1 by blast


It's not clear to me how "It is possible to do this [i.e., apply conjI twice] within an Isar proof," but perhaps I don't now need to know.






share|improve this answer




















  • 1





    "It is possible to do this [i.e., apply conjI twice] within an Isar proof,". It is possible to use a comma-separated list of methods (sequential composition, also see 6.4 in the reference manual) both in Isar and apply scripts, e.g lemma "True" proof fix A B :: "'a ⇒ bool" have "∀x y. A x ∧ B y ⟶ B y ∧ A x" by (rule allI, rule allI, simp) ....

    – xanonec
    yesterday






  • 1





    Also, there is nothing that could prevent one from using apply style script inside a proof block. To the best of my knowledge, it is not considered to be a very good practice, but it is not considered to be a particularly bad practice either (see "Do not switch from apply into proof in the middle of a proof" in https://proofcraft.org/blog/isabelle-style.html).

    – xanonec
    yesterday














0












0








0







According to @xanonec:




I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.




It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.




According to @John:
The syntax for this process that actually worked was this:



 proposition four_points_sufficient: "affine_plane plmeets"
unfolding affine_plane_def
apply (intro conjI)
subgoal using four_points_a1 by blast


It's not clear to me how "It is possible to do this [i.e., apply conjI twice] within an Isar proof," but perhaps I don't now need to know.






share|improve this answer















According to @xanonec:




I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.




It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.




According to @John:
The syntax for this process that actually worked was this:



 proposition four_points_sufficient: "affine_plane plmeets"
unfolding affine_plane_def
apply (intro conjI)
subgoal using four_points_a1 by blast


It's not clear to me how "It is possible to do this [i.e., apply conjI twice] within an Isar proof," but perhaps I don't now need to know.







share|improve this answer














share|improve this answer



share|improve this answer








answered 2 days ago


























community wiki





John








  • 1





    "It is possible to do this [i.e., apply conjI twice] within an Isar proof,". It is possible to use a comma-separated list of methods (sequential composition, also see 6.4 in the reference manual) both in Isar and apply scripts, e.g lemma "True" proof fix A B :: "'a ⇒ bool" have "∀x y. A x ∧ B y ⟶ B y ∧ A x" by (rule allI, rule allI, simp) ....

    – xanonec
    yesterday






  • 1





    Also, there is nothing that could prevent one from using apply style script inside a proof block. To the best of my knowledge, it is not considered to be a very good practice, but it is not considered to be a particularly bad practice either (see "Do not switch from apply into proof in the middle of a proof" in https://proofcraft.org/blog/isabelle-style.html).

    – xanonec
    yesterday













  • 1





    "It is possible to do this [i.e., apply conjI twice] within an Isar proof,". It is possible to use a comma-separated list of methods (sequential composition, also see 6.4 in the reference manual) both in Isar and apply scripts, e.g lemma "True" proof fix A B :: "'a ⇒ bool" have "∀x y. A x ∧ B y ⟶ B y ∧ A x" by (rule allI, rule allI, simp) ....

    – xanonec
    yesterday






  • 1





    Also, there is nothing that could prevent one from using apply style script inside a proof block. To the best of my knowledge, it is not considered to be a very good practice, but it is not considered to be a particularly bad practice either (see "Do not switch from apply into proof in the middle of a proof" in https://proofcraft.org/blog/isabelle-style.html).

    – xanonec
    yesterday








1




1





"It is possible to do this [i.e., apply conjI twice] within an Isar proof,". It is possible to use a comma-separated list of methods (sequential composition, also see 6.4 in the reference manual) both in Isar and apply scripts, e.g lemma "True" proof fix A B :: "'a ⇒ bool" have "∀x y. A x ∧ B y ⟶ B y ∧ A x" by (rule allI, rule allI, simp) ....

– xanonec
yesterday





"It is possible to do this [i.e., apply conjI twice] within an Isar proof,". It is possible to use a comma-separated list of methods (sequential composition, also see 6.4 in the reference manual) both in Isar and apply scripts, e.g lemma "True" proof fix A B :: "'a ⇒ bool" have "∀x y. A x ∧ B y ⟶ B y ∧ A x" by (rule allI, rule allI, simp) ....

– xanonec
yesterday




1




1





Also, there is nothing that could prevent one from using apply style script inside a proof block. To the best of my knowledge, it is not considered to be a very good practice, but it is not considered to be a particularly bad practice either (see "Do not switch from apply into proof in the middle of a proof" in https://proofcraft.org/blog/isabelle-style.html).

– xanonec
yesterday






Also, there is nothing that could prevent one from using apply style script inside a proof block. To the best of my knowledge, it is not considered to be a very good practice, but it is not considered to be a particularly bad practice either (see "Do not switch from apply into proof in the middle of a proof" in https://proofcraft.org/blog/isabelle-style.html).

– xanonec
yesterday




















draft saved

draft discarded
















































Thanks for contributing an answer to Stack Overflow!


  • 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.

To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55028088%2fisar-proof-of-conjunction%23new-answer', 'question_page');

);

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







Popular posts from this blog

Thal And Out Agency railway station See also References External links Navigation menuOfficial Web Site of Pakistan RailwaysArchivedOfficial Web Site of Pakistan Railwayseeexpanding ite

Posting a File and Associated Data to a RESTful WebService preferably as JSON2019 Community Moderator ElectionREST URI convention - Singular or plural name of resource while creating itUploading both data and files in one form using Ajax?How do I upload a file with metadata using a REST web service?REST API - file (ie images) processing - best practicesSpring: JSON data and file in the same requestREST API Put large dataPosting a file and JSON data to Spring rest serviceRESTful Web Service Upload/Download Large Data With JSONREST API Design sending JSON data and a file to the api in same requestPost JSONArray to REST servicePUT vs. POST in RESTParsing values from a JSON file?JavaScript/jQuery to download file via POST with JSON dataHow do I upload a file with metadata using a REST web service?How to POST JSON data with Curl from Terminal/Commandline to Test Spring REST?Separate REST JSON API server and client?what's the correct way to send a file from REST web service to client?How do I write JSON data to a file?What is “406-Not Acceptable Response” in HTTP?REST API - file (ie images) processing - best practices

How can I change the color of pagination dots of UIPageControl?How to change UIPageControl dotsIs there a way to change page indicator dots colorCustomize dot with image of UIPageControl at index 0 of UIPageControlNo visible @interface for 'NSObject<PageControlDelegate>' declares the selector 'pageControlPageDidChange:'How to change the color of pagination dots in UIPageControl with a different color per pagepagecontrol indicator custom image instead of DefaultChanging the colour of UIPageControl dots in MonoTouchpagecontrol selectable page visibility color?Alternative way to load ViewControllers on a UIPageControlHow to set only layer.border-color for UIpage control dots in swiftHow can I develop for iPhone using a Windows development machine?How to change the name of an iOS app?UITableView - change section header coloruipagecontrol indicator(dot)issueCustom UIPageControl dots color not changingchange the interspace between UIPageControl dotsHow to change Status Bar text color in iOSHow can I change image tintColor in iOS and WatchKitUIPageControl dots with larger space in between each dotsHow to change the color of pagination dots in UIPageControl with a different color per page