Propositional logic, valid entailment

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











up vote
1
down vote

favorite












I have an old exam-question but don't know how to solve
this type of problems. So If someone could give me a hint it would be much appreciated.



Is the following entailment valid?



$$
(p rightarrow q) rightarrow r, neg rland neg s, (q rightarrow p) lor t, t rightarrow (r lor p) vdash t
$$



If I assume that the right handside is false$(neg t)$ and show that the left handside holds, then the entailment is not valid. But how?




logic propositional-calculus




share|cite|improve this question




















  • Have you made any progress?
    – Graham Kemp
    Aug 21 at 2:14














up vote
1
down vote

favorite












I have an old exam-question but don't know how to solve
this type of problems. So If someone could give me a hint it would be much appreciated.



Is the following entailment valid?



$$
(p rightarrow q) rightarrow r, neg rland neg s, (q rightarrow p) lor t, t rightarrow (r lor p) vdash t
$$



If I assume that the right handside is false$(neg t)$ and show that the left handside holds, then the entailment is not valid. But how?




logic propositional-calculus




share|cite|improve this question




















  • Have you made any progress?
    – Graham Kemp
    Aug 21 at 2:14












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have an old exam-question but don't know how to solve
this type of problems. So If someone could give me a hint it would be much appreciated.



Is the following entailment valid?



$$
(p rightarrow q) rightarrow r, neg rland neg s, (q rightarrow p) lor t, t rightarrow (r lor p) vdash t
$$



If I assume that the right handside is false$(neg t)$ and show that the left handside holds, then the entailment is not valid. But how?




logic propositional-calculus




share|cite|improve this question












I have an old exam-question but don't know how to solve
this type of problems. So If someone could give me a hint it would be much appreciated.



Is the following entailment valid?



$$
(p rightarrow q) rightarrow r, neg rland neg s, (q rightarrow p) lor t, t rightarrow (r lor p) vdash t
$$



If I assume that the right handside is false$(neg t)$ and show that the left handside holds, then the entailment is not valid. But how?





logic propositional-calculus





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 19 at 8:27









Lars Logik

111




111











  • Have you made any progress?
    – Graham Kemp
    Aug 21 at 2:14
















  • Have you made any progress?
    – Graham Kemp
    Aug 21 at 2:14















Have you made any progress?
– Graham Kemp
Aug 21 at 2:14




Have you made any progress?
– Graham Kemp
Aug 21 at 2:14










1 Answer
1






active

oldest

votes

















up vote
1
down vote













If all of the premises and the negation of the conclusion can be satisfied, then the conclusion is not a logical entailment of the premises.



$$(p rightarrow q) rightarrow r, neg rland neg s, (q rightarrow p) lor t, t rightarrow (r lor p), neg t$$



Clearly, the negation of the conclusion may be satisfied when $t$ is false. That also satisfies the fourth premise, but the third premise would then only be satisfied when $q$ implies $p$.



$$(p rightarrow q) rightarrow r, neg rland neg s, (q rightarrow p) lor bot, bot rightarrow (r lor p), neg bot$$



Carry on..






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: 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%2f2887490%2fpropositional-logic-valid-entailment%23new-answer', 'question_page');

    );

    Post as a guest

















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote













    If all of the premises and the negation of the conclusion can be satisfied, then the conclusion is not a logical entailment of the premises.



    $$(p rightarrow q) rightarrow r, neg rland neg s, (q rightarrow p) lor t, t rightarrow (r lor p), neg t$$



    Clearly, the negation of the conclusion may be satisfied when $t$ is false. That also satisfies the fourth premise, but the third premise would then only be satisfied when $q$ implies $p$.



    $$(p rightarrow q) rightarrow r, neg rland neg s, (q rightarrow p) lor bot, bot rightarrow (r lor p), neg bot$$



    Carry on..






    share|cite|improve this answer
























      up vote
      1
      down vote













      If all of the premises and the negation of the conclusion can be satisfied, then the conclusion is not a logical entailment of the premises.



      $$(p rightarrow q) rightarrow r, neg rland neg s, (q rightarrow p) lor t, t rightarrow (r lor p), neg t$$



      Clearly, the negation of the conclusion may be satisfied when $t$ is false. That also satisfies the fourth premise, but the third premise would then only be satisfied when $q$ implies $p$.



      $$(p rightarrow q) rightarrow r, neg rland neg s, (q rightarrow p) lor bot, bot rightarrow (r lor p), neg bot$$



      Carry on..






      share|cite|improve this answer






















        up vote
        1
        down vote










        up vote
        1
        down vote









        If all of the premises and the negation of the conclusion can be satisfied, then the conclusion is not a logical entailment of the premises.



        $$(p rightarrow q) rightarrow r, neg rland neg s, (q rightarrow p) lor t, t rightarrow (r lor p), neg t$$



        Clearly, the negation of the conclusion may be satisfied when $t$ is false. That also satisfies the fourth premise, but the third premise would then only be satisfied when $q$ implies $p$.



        $$(p rightarrow q) rightarrow r, neg rland neg s, (q rightarrow p) lor bot, bot rightarrow (r lor p), neg bot$$



        Carry on..






        share|cite|improve this answer












        If all of the premises and the negation of the conclusion can be satisfied, then the conclusion is not a logical entailment of the premises.



        $$(p rightarrow q) rightarrow r, neg rland neg s, (q rightarrow p) lor t, t rightarrow (r lor p), neg t$$



        Clearly, the negation of the conclusion may be satisfied when $t$ is false. That also satisfies the fourth premise, but the third premise would then only be satisfied when $q$ implies $p$.



        $$(p rightarrow q) rightarrow r, neg rland neg s, (q rightarrow p) lor bot, bot rightarrow (r lor p), neg bot$$



        Carry on..







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 19 at 8:50









        Graham Kemp

        80.5k43275




        80.5k43275






















             

            draft saved


            draft discarded


























             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2887490%2fpropositional-logic-valid-entailment%23new-answer', 'question_page');

            );

            Post as a guest
































































            -

            這個網誌中的熱門文章

            How to combine Bézier curves to a surface?

            圖片
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
            閱讀完整內容

            Mutual Information Always Non-negative

            圖片
            Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
            閱讀完整內容

            Why am i infinitely getting the same tweet with the Twitter Search API?

            圖片
            up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia...
            閱讀完整內容