Prove (¬(∀𝑥(¬𝑃(𝑥)))) ⊢ (∃𝑥 𝑃(𝑥)) by Natural Deduction

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












I want to prove (¬(∀𝑥(¬𝑃(𝑥)))) ⊢ (∃𝑥𝑃(𝑥)) using only the basic rules of the Natural Deduction system for propositional logic and predicate logic.



I am not sure how to get rid of the negation before the universal quantifier.



How should I prove this?






logic first-order-logic predicate-logic quantifiers natural-deduction







share|cite|improve this question

























    up vote
    1
    down vote

    favorite












    I want to prove (¬(∀𝑥(¬𝑃(𝑥)))) ⊢ (∃𝑥𝑃(𝑥)) using only the basic rules of the Natural Deduction system for propositional logic and predicate logic.



    I am not sure how to get rid of the negation before the universal quantifier.



    How should I prove this?






    logic first-order-logic predicate-logic quantifiers natural-deduction







    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I want to prove (¬(∀𝑥(¬𝑃(𝑥)))) ⊢ (∃𝑥𝑃(𝑥)) using only the basic rules of the Natural Deduction system for propositional logic and predicate logic.



      I am not sure how to get rid of the negation before the universal quantifier.



      How should I prove this?






      logic first-order-logic predicate-logic quantifiers natural-deduction







      share|cite|improve this question













      I want to prove (¬(∀𝑥(¬𝑃(𝑥)))) ⊢ (∃𝑥𝑃(𝑥)) using only the basic rules of the Natural Deduction system for propositional logic and predicate logic.



      I am not sure how to get rid of the negation before the universal quantifier.



      How should I prove this?






      logic first-order-logic predicate-logic quantifiers natural-deduction




      logic first-order-logic predicate-logic quantifiers natural-deduction






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 3 hours ago









      Mathrocks

      82




      82




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          3
          down vote













          If in doubt, try reductio ....



          So after the initial premiss, assume $negexists xPx$



          Now what?



          You'll have to make another assumption to get anywhere ....



          So suppose $Pa$ (the obvious thing ...why??)



          Then you can infer $exists xPx$, contradiction!



          So you can infer $neg Pa$



          And that gives you $forall xneg Px$



          And now the end is in sight, because this contradicts the initial premiss ...



          Join up the dots, and finish the proof.






          share|cite|improve this answer



























            up vote
            1
            down vote













            Natural deduction rules are all about introduction and eliminations of connectives and quantifiers.   Hence the rules' names.   Writing a natural deduction proof basically consists of deciding what needs to be introduced or eliminated and when.   The rules themselves tell you how.




            I am not sure how to get rid of the negation before the universal quantifier.




            Okay, obviously, eliminating the negation will be done using the rule of negation elimination.   That was easy!



            Well, the hard bit of this is arranging to have what the rule requires.   Although it is not that hard.



            It requires two contradicting statements and we have only one premise in the base context.   An assumption must be raised!   So what do we assume to derive a contradiction of the premise?   Also what do we do with that contradiction?



            Well, if the premise actually does entails the conclusion, assuming the contrary should derive a contradiction.   So let us assume that: $negexists x~P(x)$ , and aim to derive this: $forall x~lnot P(x)$.   This also tells use what we're doing with that contradiction.   We will be using a proof by reduction to absurdity — discharging the assumption by introducing a negation, and then eliminating the double negation that results to derive the desired conclusion.



            Okay, now, to get that contradiction you need to introduce a universal quantifier.   Continue thinking in this manner and the proof almost writes itself.



            $$deffitch#1#2quadbeginarray#1\hline#2endarrayfitchnegforall x~neg P(x)qquad~~~textsfPremisefitchnegexists x~P(x)qquadtextsfAssumption~vdots\forall x~neg P(x)qquadtextsfUniversal Introduction\botqquadqquadquad~textsfNegation Elimination\negnegexists x~P(x)qquadquadtextsfNegation Introduction\exists x~P(x)qquadqquadtextsfDouble Negation Elimination$$






            share|cite|improve this answer




















            • That is a good proof outline.
              – DanielV
              14 mins ago










            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',
            convertImagesToLinks: true,
            noModals: false,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            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%2f2975472%2fprove-%25c2%25ac%25e2%2588%2580%25c2%25ac-%25e2%258a%25a2-%25e2%2588%2583-by-natural-deduction%23new-answer', 'question_page');

            );

            Post as a guest

















            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote













            If in doubt, try reductio ....



            So after the initial premiss, assume $negexists xPx$



            Now what?



            You'll have to make another assumption to get anywhere ....



            So suppose $Pa$ (the obvious thing ...why??)



            Then you can infer $exists xPx$, contradiction!



            So you can infer $neg Pa$



            And that gives you $forall xneg Px$



            And now the end is in sight, because this contradicts the initial premiss ...



            Join up the dots, and finish the proof.






            share|cite|improve this answer
























              up vote
              3
              down vote













              If in doubt, try reductio ....



              So after the initial premiss, assume $negexists xPx$



              Now what?



              You'll have to make another assumption to get anywhere ....



              So suppose $Pa$ (the obvious thing ...why??)



              Then you can infer $exists xPx$, contradiction!



              So you can infer $neg Pa$



              And that gives you $forall xneg Px$



              And now the end is in sight, because this contradicts the initial premiss ...



              Join up the dots, and finish the proof.






              share|cite|improve this answer






















                up vote
                3
                down vote










                up vote
                3
                down vote









                If in doubt, try reductio ....



                So after the initial premiss, assume $negexists xPx$



                Now what?



                You'll have to make another assumption to get anywhere ....



                So suppose $Pa$ (the obvious thing ...why??)



                Then you can infer $exists xPx$, contradiction!



                So you can infer $neg Pa$



                And that gives you $forall xneg Px$



                And now the end is in sight, because this contradicts the initial premiss ...



                Join up the dots, and finish the proof.






                share|cite|improve this answer












                If in doubt, try reductio ....



                So after the initial premiss, assume $negexists xPx$



                Now what?



                You'll have to make another assumption to get anywhere ....



                So suppose $Pa$ (the obvious thing ...why??)



                Then you can infer $exists xPx$, contradiction!



                So you can infer $neg Pa$



                And that gives you $forall xneg Px$



                And now the end is in sight, because this contradicts the initial premiss ...



                Join up the dots, and finish the proof.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 3 hours ago









                Peter Smith

                39.7k339118




                39.7k339118




















                    up vote
                    1
                    down vote













                    Natural deduction rules are all about introduction and eliminations of connectives and quantifiers.   Hence the rules' names.   Writing a natural deduction proof basically consists of deciding what needs to be introduced or eliminated and when.   The rules themselves tell you how.




                    I am not sure how to get rid of the negation before the universal quantifier.




                    Okay, obviously, eliminating the negation will be done using the rule of negation elimination.   That was easy!



                    Well, the hard bit of this is arranging to have what the rule requires.   Although it is not that hard.



                    It requires two contradicting statements and we have only one premise in the base context.   An assumption must be raised!   So what do we assume to derive a contradiction of the premise?   Also what do we do with that contradiction?



                    Well, if the premise actually does entails the conclusion, assuming the contrary should derive a contradiction.   So let us assume that: $negexists x~P(x)$ , and aim to derive this: $forall x~lnot P(x)$.   This also tells use what we're doing with that contradiction.   We will be using a proof by reduction to absurdity — discharging the assumption by introducing a negation, and then eliminating the double negation that results to derive the desired conclusion.



                    Okay, now, to get that contradiction you need to introduce a universal quantifier.   Continue thinking in this manner and the proof almost writes itself.



                    $$deffitch#1#2quadbeginarray#1\hline#2endarrayfitchnegforall x~neg P(x)qquad~~~textsfPremisefitchnegexists x~P(x)qquadtextsfAssumption~vdots\forall x~neg P(x)qquadtextsfUniversal Introduction\botqquadqquadquad~textsfNegation Elimination\negnegexists x~P(x)qquadquadtextsfNegation Introduction\exists x~P(x)qquadqquadtextsfDouble Negation Elimination$$






                    share|cite|improve this answer




















                    • That is a good proof outline.
                      – DanielV
                      14 mins ago














                    up vote
                    1
                    down vote













                    Natural deduction rules are all about introduction and eliminations of connectives and quantifiers.   Hence the rules' names.   Writing a natural deduction proof basically consists of deciding what needs to be introduced or eliminated and when.   The rules themselves tell you how.




                    I am not sure how to get rid of the negation before the universal quantifier.




                    Okay, obviously, eliminating the negation will be done using the rule of negation elimination.   That was easy!



                    Well, the hard bit of this is arranging to have what the rule requires.   Although it is not that hard.



                    It requires two contradicting statements and we have only one premise in the base context.   An assumption must be raised!   So what do we assume to derive a contradiction of the premise?   Also what do we do with that contradiction?



                    Well, if the premise actually does entails the conclusion, assuming the contrary should derive a contradiction.   So let us assume that: $negexists x~P(x)$ , and aim to derive this: $forall x~lnot P(x)$.   This also tells use what we're doing with that contradiction.   We will be using a proof by reduction to absurdity — discharging the assumption by introducing a negation, and then eliminating the double negation that results to derive the desired conclusion.



                    Okay, now, to get that contradiction you need to introduce a universal quantifier.   Continue thinking in this manner and the proof almost writes itself.



                    $$deffitch#1#2quadbeginarray#1\hline#2endarrayfitchnegforall x~neg P(x)qquad~~~textsfPremisefitchnegexists x~P(x)qquadtextsfAssumption~vdots\forall x~neg P(x)qquadtextsfUniversal Introduction\botqquadqquadquad~textsfNegation Elimination\negnegexists x~P(x)qquadquadtextsfNegation Introduction\exists x~P(x)qquadqquadtextsfDouble Negation Elimination$$






                    share|cite|improve this answer




















                    • That is a good proof outline.
                      – DanielV
                      14 mins ago












                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    Natural deduction rules are all about introduction and eliminations of connectives and quantifiers.   Hence the rules' names.   Writing a natural deduction proof basically consists of deciding what needs to be introduced or eliminated and when.   The rules themselves tell you how.




                    I am not sure how to get rid of the negation before the universal quantifier.




                    Okay, obviously, eliminating the negation will be done using the rule of negation elimination.   That was easy!



                    Well, the hard bit of this is arranging to have what the rule requires.   Although it is not that hard.



                    It requires two contradicting statements and we have only one premise in the base context.   An assumption must be raised!   So what do we assume to derive a contradiction of the premise?   Also what do we do with that contradiction?



                    Well, if the premise actually does entails the conclusion, assuming the contrary should derive a contradiction.   So let us assume that: $negexists x~P(x)$ , and aim to derive this: $forall x~lnot P(x)$.   This also tells use what we're doing with that contradiction.   We will be using a proof by reduction to absurdity — discharging the assumption by introducing a negation, and then eliminating the double negation that results to derive the desired conclusion.



                    Okay, now, to get that contradiction you need to introduce a universal quantifier.   Continue thinking in this manner and the proof almost writes itself.



                    $$deffitch#1#2quadbeginarray#1\hline#2endarrayfitchnegforall x~neg P(x)qquad~~~textsfPremisefitchnegexists x~P(x)qquadtextsfAssumption~vdots\forall x~neg P(x)qquadtextsfUniversal Introduction\botqquadqquadquad~textsfNegation Elimination\negnegexists x~P(x)qquadquadtextsfNegation Introduction\exists x~P(x)qquadqquadtextsfDouble Negation Elimination$$






                    share|cite|improve this answer












                    Natural deduction rules are all about introduction and eliminations of connectives and quantifiers.   Hence the rules' names.   Writing a natural deduction proof basically consists of deciding what needs to be introduced or eliminated and when.   The rules themselves tell you how.




                    I am not sure how to get rid of the negation before the universal quantifier.




                    Okay, obviously, eliminating the negation will be done using the rule of negation elimination.   That was easy!



                    Well, the hard bit of this is arranging to have what the rule requires.   Although it is not that hard.



                    It requires two contradicting statements and we have only one premise in the base context.   An assumption must be raised!   So what do we assume to derive a contradiction of the premise?   Also what do we do with that contradiction?



                    Well, if the premise actually does entails the conclusion, assuming the contrary should derive a contradiction.   So let us assume that: $negexists x~P(x)$ , and aim to derive this: $forall x~lnot P(x)$.   This also tells use what we're doing with that contradiction.   We will be using a proof by reduction to absurdity — discharging the assumption by introducing a negation, and then eliminating the double negation that results to derive the desired conclusion.



                    Okay, now, to get that contradiction you need to introduce a universal quantifier.   Continue thinking in this manner and the proof almost writes itself.



                    $$deffitch#1#2quadbeginarray#1\hline#2endarrayfitchnegforall x~neg P(x)qquad~~~textsfPremisefitchnegexists x~P(x)qquadtextsfAssumption~vdots\forall x~neg P(x)qquadtextsfUniversal Introduction\botqquadqquadquad~textsfNegation Elimination\negnegexists x~P(x)qquadquadtextsfNegation Introduction\exists x~P(x)qquadqquadtextsfDouble Negation Elimination$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    Graham Kemp

                    82.8k43378




                    82.8k43378











                    • That is a good proof outline.
                      – DanielV
                      14 mins ago
















                    • That is a good proof outline.
                      – DanielV
                      14 mins ago















                    That is a good proof outline.
                    – DanielV
                    14 mins ago




                    That is a good proof outline.
                    – DanielV
                    14 mins ago

















                     

                    draft saved


                    draft discarded















































                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2975472%2fprove-%25c2%25ac%25e2%2588%2580%25c2%25ac-%25e2%258a%25a2-%25e2%2588%2583-by-natural-deduction%23new-answer', 'question_page');

                    );

                    Post as a guest
































































                    Comments

                    Post a Comment

                    Popular posts from this blog

                    What does second last employer means? [closed]

                    Image
                    Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
                    Read more

                    Installing NextGIS Connect into QGIS 3?

                    Image
                    Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugins nextgis share | improve this question edited 3 hours ago PolyGeo ♦ 51.9k 17 77 233 asked 4 hours ago IanGS 16 4 New contributor IanGS is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugin...
                    Read more

                    One-line joke

                    A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
                    Read more