Predicate Logic Resolution












2












$begingroup$


I'm trying to use resolution to prove that:



$forall x(A(x) lor B(x)) land neg B(a) models A(a)$



My attempt was to try resolve $neg B(a)$ and $B(x)$ by forcing the substitution of x for a. This gives



$forall aA(a) models A(a)$ which is true.



Is this correct? I am unsure as I am still trying to understand resolution in predicate logic.



Thanks for any help!










share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I'm trying to use resolution to prove that:



    $forall x(A(x) lor B(x)) land neg B(a) models A(a)$



    My attempt was to try resolve $neg B(a)$ and $B(x)$ by forcing the substitution of x for a. This gives



    $forall aA(a) models A(a)$ which is true.



    Is this correct? I am unsure as I am still trying to understand resolution in predicate logic.



    Thanks for any help!










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I'm trying to use resolution to prove that:



      $forall x(A(x) lor B(x)) land neg B(a) models A(a)$



      My attempt was to try resolve $neg B(a)$ and $B(x)$ by forcing the substitution of x for a. This gives



      $forall aA(a) models A(a)$ which is true.



      Is this correct? I am unsure as I am still trying to understand resolution in predicate logic.



      Thanks for any help!










      share|cite|improve this question









      $endgroup$




      I'm trying to use resolution to prove that:



      $forall x(A(x) lor B(x)) land neg B(a) models A(a)$



      My attempt was to try resolve $neg B(a)$ and $B(x)$ by forcing the substitution of x for a. This gives



      $forall aA(a) models A(a)$ which is true.



      Is this correct? I am unsure as I am still trying to understand resolution in predicate logic.



      Thanks for any help!







      logic predicate-logic






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 18 '18 at 0:32









      martinhynesonemartinhynesone

      367




      367






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          To do resolution, you need to negate the conclusion, put that into clauses, add that to the clauses from the premises, and then derive the empty clause.



          Thus, you get the clause ${neg A(a) }$ from the negation of the conclusion, and clauses ${ A(x), B(x) }$ and ${ neg B(a) }$ from the premises.



          Now you can resolve the latter two clauses by setting $x=a$, resulting in the clause ${ A(a) }$.



          And that clause resolves with the ${ neg A(a) }$ clause, resulting in the empty clause ${ }$



          That ends the resolution, but to explain why that proves the argument to be valid: the empty represents a contradiction, and so what you have shown is that if you assume th conclusion is not true, you get a contradiction which, by Proof by Contradiction, shows that the conclusion has to be true given the turht of the premises ... meaning the argument is valid.






          share|cite|improve this answer









          $endgroup$













            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%2f3044646%2fpredicate-logic-resolution%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









            3












            $begingroup$

            To do resolution, you need to negate the conclusion, put that into clauses, add that to the clauses from the premises, and then derive the empty clause.



            Thus, you get the clause ${neg A(a) }$ from the negation of the conclusion, and clauses ${ A(x), B(x) }$ and ${ neg B(a) }$ from the premises.



            Now you can resolve the latter two clauses by setting $x=a$, resulting in the clause ${ A(a) }$.



            And that clause resolves with the ${ neg A(a) }$ clause, resulting in the empty clause ${ }$



            That ends the resolution, but to explain why that proves the argument to be valid: the empty represents a contradiction, and so what you have shown is that if you assume th conclusion is not true, you get a contradiction which, by Proof by Contradiction, shows that the conclusion has to be true given the turht of the premises ... meaning the argument is valid.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              To do resolution, you need to negate the conclusion, put that into clauses, add that to the clauses from the premises, and then derive the empty clause.



              Thus, you get the clause ${neg A(a) }$ from the negation of the conclusion, and clauses ${ A(x), B(x) }$ and ${ neg B(a) }$ from the premises.



              Now you can resolve the latter two clauses by setting $x=a$, resulting in the clause ${ A(a) }$.



              And that clause resolves with the ${ neg A(a) }$ clause, resulting in the empty clause ${ }$



              That ends the resolution, but to explain why that proves the argument to be valid: the empty represents a contradiction, and so what you have shown is that if you assume th conclusion is not true, you get a contradiction which, by Proof by Contradiction, shows that the conclusion has to be true given the turht of the premises ... meaning the argument is valid.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                To do resolution, you need to negate the conclusion, put that into clauses, add that to the clauses from the premises, and then derive the empty clause.



                Thus, you get the clause ${neg A(a) }$ from the negation of the conclusion, and clauses ${ A(x), B(x) }$ and ${ neg B(a) }$ from the premises.



                Now you can resolve the latter two clauses by setting $x=a$, resulting in the clause ${ A(a) }$.



                And that clause resolves with the ${ neg A(a) }$ clause, resulting in the empty clause ${ }$



                That ends the resolution, but to explain why that proves the argument to be valid: the empty represents a contradiction, and so what you have shown is that if you assume th conclusion is not true, you get a contradiction which, by Proof by Contradiction, shows that the conclusion has to be true given the turht of the premises ... meaning the argument is valid.






                share|cite|improve this answer









                $endgroup$



                To do resolution, you need to negate the conclusion, put that into clauses, add that to the clauses from the premises, and then derive the empty clause.



                Thus, you get the clause ${neg A(a) }$ from the negation of the conclusion, and clauses ${ A(x), B(x) }$ and ${ neg B(a) }$ from the premises.



                Now you can resolve the latter two clauses by setting $x=a$, resulting in the clause ${ A(a) }$.



                And that clause resolves with the ${ neg A(a) }$ clause, resulting in the empty clause ${ }$



                That ends the resolution, but to explain why that proves the argument to be valid: the empty represents a contradiction, and so what you have shown is that if you assume th conclusion is not true, you get a contradiction which, by Proof by Contradiction, shows that the conclusion has to be true given the turht of the premises ... meaning the argument is valid.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 18 '18 at 1:10









                Bram28Bram28

                63.9k44793




                63.9k44793






























                    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%2f3044646%2fpredicate-logic-resolution%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

                    Plaza Victoria

                    Puebla de Zaragoza

                    Musa