Using inference to prove that $(pwedge q)vee (pwedge r) to (q vee r)$.












1












$begingroup$


Only using rules of inferences to prove that $(pwedge q)vee (pwedge r) to (q vee r)$



I tried to solve it with this:




  1. (p⋀q)⋁(p⋀r) (premise)

  2. (p⋀q) (Assumption)

  3. p (⋀-elimination)

  4. p⋀r (Assumption)

  5. p (⋀-elimination)

  6. p (⋁-elimination 3,5)


But this seems wrong. Does anyone else have a clue on which rules to apply for this? Thank you.






discrete-mathematics logic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What rules of inference are you allowed to use? Impossible to answer without knowing this.
    $endgroup$
    – David
    Feb 1 at 3:42












  • $begingroup$
    Introduction rules and elimination rules
    $endgroup$
    – alice123019
    Feb 1 at 3:50










  • $begingroup$
    What have you tried? It is a fairly easy proof, so where are you having trouble?
    $endgroup$
    – Graham Kemp
    Feb 1 at 3:52










  • $begingroup$
    Do you have the distributive law?
    $endgroup$
    – vadim123
    Feb 1 at 4:10










  • $begingroup$
    Distributive law is not allowed to use in this question
    $endgroup$
    – alice123019
    Feb 1 at 4:12
















1












$begingroup$


Only using rules of inferences to prove that $(pwedge q)vee (pwedge r) to (q vee r)$



I tried to solve it with this:




  1. (p⋀q)⋁(p⋀r) (premise)

  2. (p⋀q) (Assumption)

  3. p (⋀-elimination)

  4. p⋀r (Assumption)

  5. p (⋀-elimination)

  6. p (⋁-elimination 3,5)


But this seems wrong. Does anyone else have a clue on which rules to apply for this? Thank you.






discrete-mathematics logic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What rules of inference are you allowed to use? Impossible to answer without knowing this.
    $endgroup$
    – David
    Feb 1 at 3:42












  • $begingroup$
    Introduction rules and elimination rules
    $endgroup$
    – alice123019
    Feb 1 at 3:50










  • $begingroup$
    What have you tried? It is a fairly easy proof, so where are you having trouble?
    $endgroup$
    – Graham Kemp
    Feb 1 at 3:52










  • $begingroup$
    Do you have the distributive law?
    $endgroup$
    – vadim123
    Feb 1 at 4:10










  • $begingroup$
    Distributive law is not allowed to use in this question
    $endgroup$
    – alice123019
    Feb 1 at 4:12














1












1








1





$begingroup$


Only using rules of inferences to prove that $(pwedge q)vee (pwedge r) to (q vee r)$



I tried to solve it with this:




  1. (p⋀q)⋁(p⋀r) (premise)

  2. (p⋀q) (Assumption)

  3. p (⋀-elimination)

  4. p⋀r (Assumption)

  5. p (⋀-elimination)

  6. p (⋁-elimination 3,5)


But this seems wrong. Does anyone else have a clue on which rules to apply for this? Thank you.






discrete-mathematics logic







share|cite|improve this question











$endgroup$




Only using rules of inferences to prove that $(pwedge q)vee (pwedge r) to (q vee r)$



I tried to solve it with this:




  1. (p⋀q)⋁(p⋀r) (premise)

  2. (p⋀q) (Assumption)

  3. p (⋀-elimination)

  4. p⋀r (Assumption)

  5. p (⋀-elimination)

  6. p (⋁-elimination 3,5)


But this seems wrong. Does anyone else have a clue on which rules to apply for this? Thank you.






discrete-mathematics logic




discrete-mathematics logic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 1 at 4:11







alice123019

















asked Feb 1 at 3:36









alice123019alice123019

84




84








  • 1




    $begingroup$
    What rules of inference are you allowed to use? Impossible to answer without knowing this.
    $endgroup$
    – David
    Feb 1 at 3:42












  • $begingroup$
    Introduction rules and elimination rules
    $endgroup$
    – alice123019
    Feb 1 at 3:50










  • $begingroup$
    What have you tried? It is a fairly easy proof, so where are you having trouble?
    $endgroup$
    – Graham Kemp
    Feb 1 at 3:52










  • $begingroup$
    Do you have the distributive law?
    $endgroup$
    – vadim123
    Feb 1 at 4:10










  • $begingroup$
    Distributive law is not allowed to use in this question
    $endgroup$
    – alice123019
    Feb 1 at 4:12














  • 1




    $begingroup$
    What rules of inference are you allowed to use? Impossible to answer without knowing this.
    $endgroup$
    – David
    Feb 1 at 3:42












  • $begingroup$
    Introduction rules and elimination rules
    $endgroup$
    – alice123019
    Feb 1 at 3:50










  • $begingroup$
    What have you tried? It is a fairly easy proof, so where are you having trouble?
    $endgroup$
    – Graham Kemp
    Feb 1 at 3:52










  • $begingroup$
    Do you have the distributive law?
    $endgroup$
    – vadim123
    Feb 1 at 4:10










  • $begingroup$
    Distributive law is not allowed to use in this question
    $endgroup$
    – alice123019
    Feb 1 at 4:12








1




1




$begingroup$
What rules of inference are you allowed to use? Impossible to answer without knowing this.
$endgroup$
– David
Feb 1 at 3:42






$begingroup$
What rules of inference are you allowed to use? Impossible to answer without knowing this.
$endgroup$
– David
Feb 1 at 3:42














$begingroup$
Introduction rules and elimination rules
$endgroup$
– alice123019
Feb 1 at 3:50




$begingroup$
Introduction rules and elimination rules
$endgroup$
– alice123019
Feb 1 at 3:50












$begingroup$
What have you tried? It is a fairly easy proof, so where are you having trouble?
$endgroup$
– Graham Kemp
Feb 1 at 3:52




$begingroup$
What have you tried? It is a fairly easy proof, so where are you having trouble?
$endgroup$
– Graham Kemp
Feb 1 at 3:52












$begingroup$
Do you have the distributive law?
$endgroup$
– vadim123
Feb 1 at 4:10




$begingroup$
Do you have the distributive law?
$endgroup$
– vadim123
Feb 1 at 4:10












$begingroup$
Distributive law is not allowed to use in this question
$endgroup$
– alice123019
Feb 1 at 4:12




$begingroup$
Distributive law is not allowed to use in this question
$endgroup$
– alice123019
Feb 1 at 4:12










1 Answer
1






active

oldest

votes


















0












$begingroup$

$deffitch#1#2{quadbegin{array}{|l}#1\hline #2end{array}}$




I tried to solve it with this:




  1. (p⋀q)⋁(p⋀r) → (q⋁r) (premise)

  2. (p⋀q) (Assumption)

  3. p (⋀-elimination)

  4. p⋀r (Assumption)

  5. p (⋀-elimination)




Don't ever start by premising what you seek to prove.   This is a proof with no premises.



Since you want to introduce a conditional, use a conditional proof -- assume the antecedant in order to derive the consequent.



Since that assumption is a disjunction, eliminate it with a proof by cases. In each case assume the left or right of the disjunction and aim to derive the consequent.



Since each case is a conjunction, eliminate it to the part that enables you to derive the consequent by introducing a disjunction.



Finish by introducing the conditional and you are done.



Fill in the dots.



$$fitch{}{fitch{ldots}{fitch{ldots}{ldots\ldots}\fitch{ldots}{ldots\ldots}\ldots}\((pland q)lor(pland r))to(qlor r)}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Gotcha! Thanks for the great explanation.
    $endgroup$
    – alice123019
    Feb 1 at 4:29












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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3095789%2fusing-inference-to-prove-that-p-wedge-q-vee-p-wedge-r-to-q-vee-r%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












$begingroup$

$deffitch#1#2{quadbegin{array}{|l}#1\hline #2end{array}}$




I tried to solve it with this:




  1. (p⋀q)⋁(p⋀r) → (q⋁r) (premise)

  2. (p⋀q) (Assumption)

  3. p (⋀-elimination)

  4. p⋀r (Assumption)

  5. p (⋀-elimination)




Don't ever start by premising what you seek to prove.   This is a proof with no premises.



Since you want to introduce a conditional, use a conditional proof -- assume the antecedant in order to derive the consequent.



Since that assumption is a disjunction, eliminate it with a proof by cases. In each case assume the left or right of the disjunction and aim to derive the consequent.



Since each case is a conjunction, eliminate it to the part that enables you to derive the consequent by introducing a disjunction.



Finish by introducing the conditional and you are done.



Fill in the dots.



$$fitch{}{fitch{ldots}{fitch{ldots}{ldots\ldots}\fitch{ldots}{ldots\ldots}\ldots}\((pland q)lor(pland r))to(qlor r)}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Gotcha! Thanks for the great explanation.
    $endgroup$
    – alice123019
    Feb 1 at 4:29
















0












$begingroup$

$deffitch#1#2{quadbegin{array}{|l}#1\hline #2end{array}}$




I tried to solve it with this:




  1. (p⋀q)⋁(p⋀r) → (q⋁r) (premise)

  2. (p⋀q) (Assumption)

  3. p (⋀-elimination)

  4. p⋀r (Assumption)

  5. p (⋀-elimination)




Don't ever start by premising what you seek to prove.   This is a proof with no premises.



Since you want to introduce a conditional, use a conditional proof -- assume the antecedant in order to derive the consequent.



Since that assumption is a disjunction, eliminate it with a proof by cases. In each case assume the left or right of the disjunction and aim to derive the consequent.



Since each case is a conjunction, eliminate it to the part that enables you to derive the consequent by introducing a disjunction.



Finish by introducing the conditional and you are done.



Fill in the dots.



$$fitch{}{fitch{ldots}{fitch{ldots}{ldots\ldots}\fitch{ldots}{ldots\ldots}\ldots}\((pland q)lor(pland r))to(qlor r)}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Gotcha! Thanks for the great explanation.
    $endgroup$
    – alice123019
    Feb 1 at 4:29














0












0








0





$begingroup$

$deffitch#1#2{quadbegin{array}{|l}#1\hline #2end{array}}$




I tried to solve it with this:




  1. (p⋀q)⋁(p⋀r) → (q⋁r) (premise)

  2. (p⋀q) (Assumption)

  3. p (⋀-elimination)

  4. p⋀r (Assumption)

  5. p (⋀-elimination)




Don't ever start by premising what you seek to prove.   This is a proof with no premises.



Since you want to introduce a conditional, use a conditional proof -- assume the antecedant in order to derive the consequent.



Since that assumption is a disjunction, eliminate it with a proof by cases. In each case assume the left or right of the disjunction and aim to derive the consequent.



Since each case is a conjunction, eliminate it to the part that enables you to derive the consequent by introducing a disjunction.



Finish by introducing the conditional and you are done.



Fill in the dots.



$$fitch{}{fitch{ldots}{fitch{ldots}{ldots\ldots}\fitch{ldots}{ldots\ldots}\ldots}\((pland q)lor(pland r))to(qlor r)}$$






share|cite|improve this answer









$endgroup$



$deffitch#1#2{quadbegin{array}{|l}#1\hline #2end{array}}$




I tried to solve it with this:




  1. (p⋀q)⋁(p⋀r) → (q⋁r) (premise)

  2. (p⋀q) (Assumption)

  3. p (⋀-elimination)

  4. p⋀r (Assumption)

  5. p (⋀-elimination)




Don't ever start by premising what you seek to prove.   This is a proof with no premises.



Since you want to introduce a conditional, use a conditional proof -- assume the antecedant in order to derive the consequent.



Since that assumption is a disjunction, eliminate it with a proof by cases. In each case assume the left or right of the disjunction and aim to derive the consequent.



Since each case is a conjunction, eliminate it to the part that enables you to derive the consequent by introducing a disjunction.



Finish by introducing the conditional and you are done.



Fill in the dots.



$$fitch{}{fitch{ldots}{fitch{ldots}{ldots\ldots}\fitch{ldots}{ldots\ldots}\ldots}\((pland q)lor(pland r))to(qlor r)}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Feb 1 at 4:17









Graham KempGraham Kemp

87.8k43578




87.8k43578












  • $begingroup$
    Gotcha! Thanks for the great explanation.
    $endgroup$
    – alice123019
    Feb 1 at 4:29


















  • $begingroup$
    Gotcha! Thanks for the great explanation.
    $endgroup$
    – alice123019
    Feb 1 at 4:29
















$begingroup$
Gotcha! Thanks for the great explanation.
$endgroup$
– alice123019
Feb 1 at 4:29




$begingroup$
Gotcha! Thanks for the great explanation.
$endgroup$
– alice123019
Feb 1 at 4:29


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3095789%2fusing-inference-to-prove-that-p-wedge-q-vee-p-wedge-r-to-q-vee-r%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

MongoDB - Not Authorized To Execute Command

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
Read more

How to fix TextFormField cause rebuild widget in Flutter

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
Read more

in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

Image
1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
Read more