If Φ⊢¬(ϕ→ψ) , show that Φ,¬ϕ and Φ, ψ are both inconsistent.











up vote
-1
down vote

favorite












So far I have shown that Φ, ψ is inconsistent:



If Φ⊢¬(ϕ→ψ) then Φ, ψ⊢¬(ϕ→ψ)



By the axiom ψ→ϕ→ψ and Modus Ponens, Φ, ψ⊢ϕ→ψ.



So Φ, ψ is inconsistent.



Could anyone help me to prove that Φ,¬ϕ is inconsistent?










share|cite|improve this question
























  • I do not get the gist of your question: what is $Phi$, $phi$ and $psi$? Which conditions hold for them? $Phi,psivdashlnot(phitopsi)$ is the only condition you have assumed?
    – Hanul Jeon
    Nov 18 at 8:06










  • I missed a step here. If Φ⊢¬(ϕ→ψ) then Φ, ψ⊢¬(ϕ→ψ).
    – mdryizk
    Nov 18 at 8:25










  • Hint: $Phivdash philand lnotpsi$.
    – Hanul Jeon
    Nov 18 at 8:26










  • Thank you for the hint. I think i know what you are trying to tell me. But i'm looking for a proof using only ¬ and →.
    – mdryizk
    Nov 18 at 8:30








  • 1




    @mdryizk If you're looking for a proof in a particular formal system you should indicate this in your question.
    – spaceisdarkgreen
    Nov 18 at 8:48

















up vote
-1
down vote

favorite












So far I have shown that Φ, ψ is inconsistent:



If Φ⊢¬(ϕ→ψ) then Φ, ψ⊢¬(ϕ→ψ)



By the axiom ψ→ϕ→ψ and Modus Ponens, Φ, ψ⊢ϕ→ψ.



So Φ, ψ is inconsistent.



Could anyone help me to prove that Φ,¬ϕ is inconsistent?










share|cite|improve this question
























  • I do not get the gist of your question: what is $Phi$, $phi$ and $psi$? Which conditions hold for them? $Phi,psivdashlnot(phitopsi)$ is the only condition you have assumed?
    – Hanul Jeon
    Nov 18 at 8:06










  • I missed a step here. If Φ⊢¬(ϕ→ψ) then Φ, ψ⊢¬(ϕ→ψ).
    – mdryizk
    Nov 18 at 8:25










  • Hint: $Phivdash philand lnotpsi$.
    – Hanul Jeon
    Nov 18 at 8:26










  • Thank you for the hint. I think i know what you are trying to tell me. But i'm looking for a proof using only ¬ and →.
    – mdryizk
    Nov 18 at 8:30








  • 1




    @mdryizk If you're looking for a proof in a particular formal system you should indicate this in your question.
    – spaceisdarkgreen
    Nov 18 at 8:48















up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











So far I have shown that Φ, ψ is inconsistent:



If Φ⊢¬(ϕ→ψ) then Φ, ψ⊢¬(ϕ→ψ)



By the axiom ψ→ϕ→ψ and Modus Ponens, Φ, ψ⊢ϕ→ψ.



So Φ, ψ is inconsistent.



Could anyone help me to prove that Φ,¬ϕ is inconsistent?










share|cite|improve this question















So far I have shown that Φ, ψ is inconsistent:



If Φ⊢¬(ϕ→ψ) then Φ, ψ⊢¬(ϕ→ψ)



By the axiom ψ→ϕ→ψ and Modus Ponens, Φ, ψ⊢ϕ→ψ.



So Φ, ψ is inconsistent.



Could anyone help me to prove that Φ,¬ϕ is inconsistent?







propositional-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 18 at 9:11

























asked Nov 18 at 8:03









mdryizk

153




153












  • I do not get the gist of your question: what is $Phi$, $phi$ and $psi$? Which conditions hold for them? $Phi,psivdashlnot(phitopsi)$ is the only condition you have assumed?
    – Hanul Jeon
    Nov 18 at 8:06










  • I missed a step here. If Φ⊢¬(ϕ→ψ) then Φ, ψ⊢¬(ϕ→ψ).
    – mdryizk
    Nov 18 at 8:25










  • Hint: $Phivdash philand lnotpsi$.
    – Hanul Jeon
    Nov 18 at 8:26










  • Thank you for the hint. I think i know what you are trying to tell me. But i'm looking for a proof using only ¬ and →.
    – mdryizk
    Nov 18 at 8:30








  • 1




    @mdryizk If you're looking for a proof in a particular formal system you should indicate this in your question.
    – spaceisdarkgreen
    Nov 18 at 8:48




















  • I do not get the gist of your question: what is $Phi$, $phi$ and $psi$? Which conditions hold for them? $Phi,psivdashlnot(phitopsi)$ is the only condition you have assumed?
    – Hanul Jeon
    Nov 18 at 8:06










  • I missed a step here. If Φ⊢¬(ϕ→ψ) then Φ, ψ⊢¬(ϕ→ψ).
    – mdryizk
    Nov 18 at 8:25










  • Hint: $Phivdash philand lnotpsi$.
    – Hanul Jeon
    Nov 18 at 8:26










  • Thank you for the hint. I think i know what you are trying to tell me. But i'm looking for a proof using only ¬ and →.
    – mdryizk
    Nov 18 at 8:30








  • 1




    @mdryizk If you're looking for a proof in a particular formal system you should indicate this in your question.
    – spaceisdarkgreen
    Nov 18 at 8:48


















I do not get the gist of your question: what is $Phi$, $phi$ and $psi$? Which conditions hold for them? $Phi,psivdashlnot(phitopsi)$ is the only condition you have assumed?
– Hanul Jeon
Nov 18 at 8:06




I do not get the gist of your question: what is $Phi$, $phi$ and $psi$? Which conditions hold for them? $Phi,psivdashlnot(phitopsi)$ is the only condition you have assumed?
– Hanul Jeon
Nov 18 at 8:06












I missed a step here. If Φ⊢¬(ϕ→ψ) then Φ, ψ⊢¬(ϕ→ψ).
– mdryizk
Nov 18 at 8:25




I missed a step here. If Φ⊢¬(ϕ→ψ) then Φ, ψ⊢¬(ϕ→ψ).
– mdryizk
Nov 18 at 8:25












Hint: $Phivdash philand lnotpsi$.
– Hanul Jeon
Nov 18 at 8:26




Hint: $Phivdash philand lnotpsi$.
– Hanul Jeon
Nov 18 at 8:26












Thank you for the hint. I think i know what you are trying to tell me. But i'm looking for a proof using only ¬ and →.
– mdryizk
Nov 18 at 8:30






Thank you for the hint. I think i know what you are trying to tell me. But i'm looking for a proof using only ¬ and →.
– mdryizk
Nov 18 at 8:30






1




1




@mdryizk If you're looking for a proof in a particular formal system you should indicate this in your question.
– spaceisdarkgreen
Nov 18 at 8:48






@mdryizk If you're looking for a proof in a particular formal system you should indicate this in your question.
– spaceisdarkgreen
Nov 18 at 8:48












1 Answer
1






active

oldest

votes

















up vote
0
down vote



accepted










From the same reasoning you used for the first part, $$ Phi,lnotphi vdash lnot psi to lnot phi.$$ By the contrapositive axiom (theorem? you haven't told us which system you are using... it seems clear that it's Hilbert style from context, but there are several commonly used versions of Axiom 3), $$ Phi,lnot phivdash (lnotpsito lnot phi)to(phito psi), $$ so by MP $$Phi,lnot phivdash phitopsi $$






share|cite|improve this answer





















    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: 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%2f3003246%2fif-%25ce%25a6%25e2%258a%25a2%25c2%25ac%25cf%2595%25e2%2586%2592%25cf%2588-show-that-%25ce%25a6-%25c2%25ac%25cf%2595-and-%25ce%25a6-%25cf%2588-are-both-inconsistent%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








    up vote
    0
    down vote



    accepted










    From the same reasoning you used for the first part, $$ Phi,lnotphi vdash lnot psi to lnot phi.$$ By the contrapositive axiom (theorem? you haven't told us which system you are using... it seems clear that it's Hilbert style from context, but there are several commonly used versions of Axiom 3), $$ Phi,lnot phivdash (lnotpsito lnot phi)to(phito psi), $$ so by MP $$Phi,lnot phivdash phitopsi $$






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      From the same reasoning you used for the first part, $$ Phi,lnotphi vdash lnot psi to lnot phi.$$ By the contrapositive axiom (theorem? you haven't told us which system you are using... it seems clear that it's Hilbert style from context, but there are several commonly used versions of Axiom 3), $$ Phi,lnot phivdash (lnotpsito lnot phi)to(phito psi), $$ so by MP $$Phi,lnot phivdash phitopsi $$






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        From the same reasoning you used for the first part, $$ Phi,lnotphi vdash lnot psi to lnot phi.$$ By the contrapositive axiom (theorem? you haven't told us which system you are using... it seems clear that it's Hilbert style from context, but there are several commonly used versions of Axiom 3), $$ Phi,lnot phivdash (lnotpsito lnot phi)to(phito psi), $$ so by MP $$Phi,lnot phivdash phitopsi $$






        share|cite|improve this answer












        From the same reasoning you used for the first part, $$ Phi,lnotphi vdash lnot psi to lnot phi.$$ By the contrapositive axiom (theorem? you haven't told us which system you are using... it seems clear that it's Hilbert style from context, but there are several commonly used versions of Axiom 3), $$ Phi,lnot phivdash (lnotpsito lnot phi)to(phito psi), $$ so by MP $$Phi,lnot phivdash phitopsi $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 18 at 9:01









        spaceisdarkgreen

        31.7k21552




        31.7k21552






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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%2fmath.stackexchange.com%2fquestions%2f3003246%2fif-%25ce%25a6%25e2%258a%25a2%25c2%25ac%25cf%2595%25e2%2586%2592%25cf%2588-show-that-%25ce%25a6-%25c2%25ac%25cf%2595-and-%25ce%25a6-%25cf%2588-are-both-inconsistent%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

            Актюбинская область

            QoS: MAC-Priority for clients behind a repeater

            AnyDesk - Fatal Program Failure