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The barbers paradox first order logic formalization

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The barbers paradox first order logic formalization


first order logic resolution unificationWhat is the relation between First Order Logic and First Order Theory?Quantified Boolean Formula vs First-order logicAbout the first order logic (valid, Unsatisfiable, Syntactically wrong)Resolution in First Order LogicExercise about First-order logicFirst Order Logic : PredicatesFirst Order Logic, First Order Logic + Recurrence and SQLReal world applications of first order logicTerminology First-Order Logic













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$begingroup$


I tried to look on the site and while I found some similar questions, I did not find the first order logic formalization of the following sentence (the basic barber's paradox), so I wanted to ask if I got the right first order logic formalization of it, and I am sure others will benefit from this thread as well.



Sentence: there exists a barber who shaves all the people that don't shave themselves.



My attempt: this complicated sentence can be made simpler by inferring that "for all of those who does not shave themselves, are shaved by the barber"



And now it seems to be easier to substitute with variables so:



$$exists x(lnot S(x,x) rightarrow S(b,x))$$



where $S(x,x)$ is shaving(verb), $x$ are the persons (who don't shave themselves), and $b$ is a shortcut for barber, so if a person does not shave himself, it can be inferred that he is shaved by the barber.






logic artificial-intelligence first-order-logic







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I tried to look on the site and while I found some similar questions, I did not find the first order logic formalization of the following sentence (the basic barber's paradox), so I wanted to ask if I got the right first order logic formalization of it, and I am sure others will benefit from this thread as well.



    Sentence: there exists a barber who shaves all the people that don't shave themselves.



    My attempt: this complicated sentence can be made simpler by inferring that "for all of those who does not shave themselves, are shaved by the barber"



    And now it seems to be easier to substitute with variables so:



    $$exists x(lnot S(x,x) rightarrow S(b,x))$$



    where $S(x,x)$ is shaving(verb), $x$ are the persons (who don't shave themselves), and $b$ is a shortcut for barber, so if a person does not shave himself, it can be inferred that he is shaved by the barber.






    logic artificial-intelligence first-order-logic







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      0



      $begingroup$


      I tried to look on the site and while I found some similar questions, I did not find the first order logic formalization of the following sentence (the basic barber's paradox), so I wanted to ask if I got the right first order logic formalization of it, and I am sure others will benefit from this thread as well.



      Sentence: there exists a barber who shaves all the people that don't shave themselves.



      My attempt: this complicated sentence can be made simpler by inferring that "for all of those who does not shave themselves, are shaved by the barber"



      And now it seems to be easier to substitute with variables so:



      $$exists x(lnot S(x,x) rightarrow S(b,x))$$



      where $S(x,x)$ is shaving(verb), $x$ are the persons (who don't shave themselves), and $b$ is a shortcut for barber, so if a person does not shave himself, it can be inferred that he is shaved by the barber.






      logic artificial-intelligence first-order-logic







      share|cite|improve this question











      $endgroup$




      I tried to look on the site and while I found some similar questions, I did not find the first order logic formalization of the following sentence (the basic barber's paradox), so I wanted to ask if I got the right first order logic formalization of it, and I am sure others will benefit from this thread as well.



      Sentence: there exists a barber who shaves all the people that don't shave themselves.



      My attempt: this complicated sentence can be made simpler by inferring that "for all of those who does not shave themselves, are shaved by the barber"



      And now it seems to be easier to substitute with variables so:



      $$exists x(lnot S(x,x) rightarrow S(b,x))$$



      where $S(x,x)$ is shaving(verb), $x$ are the persons (who don't shave themselves), and $b$ is a shortcut for barber, so if a person does not shave himself, it can be inferred that he is shaved by the barber.






      logic artificial-intelligence first-order-logic




      logic artificial-intelligence first-order-logic






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      David Richerby

      71.1k16109199




      71.1k16109199










      asked 2 hours ago









      hps13hps13

      396




      396






















          2 Answers
          2






          active

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          2












          $begingroup$

          You're not being asked to do any inference; just to express something.




          there exists a barber who shaves all the people that don't shave themselves




          translates directly as



          $$exists b, forall x,big(neg S(x,x) rightarrow S(b,x)big),.$$



          (There exists a barber such that everyone who doesn't shave themself is shaved by the barber.)



          Your version says that there exists a person who, if they don't shave themself, is shaved by the barber. That's not equivalent. For example, it's true in any world where at least one person does shave themself: just let $x$ be that person, so $neg S(x,x)$ is false, so $neg S(x,x)rightarrowtext{anything}$ is true.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            There are two issues about your formula:

            First, you use an existential quantifier for the people where there should be a univeral one, since the statement is "all people". What your formula expresses is "There is an $x$ such that if $x$ does't shave themself, the barber does." Instead, what you want to say is that "For all $x$ it holds that if $x$ doesn't shave themself, the barber does." In general, an existentially quantified implicational formula is often a sign that something is wrong.

            Second, instead of using a constant name $b$ for the barber, it would be better to directly translate the "There is" as an existential quantifier, and you could also specify that this someone is a barber.



            Whith this, the sentence becomes




            There is a thing $y$ which is a barber and such that for all people $x$ who don't shave themselves, $y$ shaves $x$.




            which translates as



            $$exists y (B(y) land forall x (neg S(x,x) to S(y,x)))$$






            share|cite|improve this answer











            $endgroup$














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              2 Answers
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              2 Answers
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              2












              $begingroup$

              You're not being asked to do any inference; just to express something.




              there exists a barber who shaves all the people that don't shave themselves




              translates directly as



              $$exists b, forall x,big(neg S(x,x) rightarrow S(b,x)big),.$$



              (There exists a barber such that everyone who doesn't shave themself is shaved by the barber.)



              Your version says that there exists a person who, if they don't shave themself, is shaved by the barber. That's not equivalent. For example, it's true in any world where at least one person does shave themself: just let $x$ be that person, so $neg S(x,x)$ is false, so $neg S(x,x)rightarrowtext{anything}$ is true.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                You're not being asked to do any inference; just to express something.




                there exists a barber who shaves all the people that don't shave themselves




                translates directly as



                $$exists b, forall x,big(neg S(x,x) rightarrow S(b,x)big),.$$



                (There exists a barber such that everyone who doesn't shave themself is shaved by the barber.)



                Your version says that there exists a person who, if they don't shave themself, is shaved by the barber. That's not equivalent. For example, it's true in any world where at least one person does shave themself: just let $x$ be that person, so $neg S(x,x)$ is false, so $neg S(x,x)rightarrowtext{anything}$ is true.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  You're not being asked to do any inference; just to express something.




                  there exists a barber who shaves all the people that don't shave themselves




                  translates directly as



                  $$exists b, forall x,big(neg S(x,x) rightarrow S(b,x)big),.$$



                  (There exists a barber such that everyone who doesn't shave themself is shaved by the barber.)



                  Your version says that there exists a person who, if they don't shave themself, is shaved by the barber. That's not equivalent. For example, it's true in any world where at least one person does shave themself: just let $x$ be that person, so $neg S(x,x)$ is false, so $neg S(x,x)rightarrowtext{anything}$ is true.






                  share|cite|improve this answer









                  $endgroup$



                  You're not being asked to do any inference; just to express something.




                  there exists a barber who shaves all the people that don't shave themselves




                  translates directly as



                  $$exists b, forall x,big(neg S(x,x) rightarrow S(b,x)big),.$$



                  (There exists a barber such that everyone who doesn't shave themself is shaved by the barber.)



                  Your version says that there exists a person who, if they don't shave themself, is shaved by the barber. That's not equivalent. For example, it's true in any world where at least one person does shave themself: just let $x$ be that person, so $neg S(x,x)$ is false, so $neg S(x,x)rightarrowtext{anything}$ is true.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  David RicherbyDavid Richerby

                  71.1k16109199




                  71.1k16109199























                      2












                      $begingroup$

                      There are two issues about your formula:

                      First, you use an existential quantifier for the people where there should be a univeral one, since the statement is "all people". What your formula expresses is "There is an $x$ such that if $x$ does't shave themself, the barber does." Instead, what you want to say is that "For all $x$ it holds that if $x$ doesn't shave themself, the barber does." In general, an existentially quantified implicational formula is often a sign that something is wrong.

                      Second, instead of using a constant name $b$ for the barber, it would be better to directly translate the "There is" as an existential quantifier, and you could also specify that this someone is a barber.



                      Whith this, the sentence becomes




                      There is a thing $y$ which is a barber and such that for all people $x$ who don't shave themselves, $y$ shaves $x$.




                      which translates as



                      $$exists y (B(y) land forall x (neg S(x,x) to S(y,x)))$$






                      share|cite|improve this answer











                      $endgroup$


















                        2












                        $begingroup$

                        There are two issues about your formula:

                        First, you use an existential quantifier for the people where there should be a univeral one, since the statement is "all people". What your formula expresses is "There is an $x$ such that if $x$ does't shave themself, the barber does." Instead, what you want to say is that "For all $x$ it holds that if $x$ doesn't shave themself, the barber does." In general, an existentially quantified implicational formula is often a sign that something is wrong.

                        Second, instead of using a constant name $b$ for the barber, it would be better to directly translate the "There is" as an existential quantifier, and you could also specify that this someone is a barber.



                        Whith this, the sentence becomes




                        There is a thing $y$ which is a barber and such that for all people $x$ who don't shave themselves, $y$ shaves $x$.




                        which translates as



                        $$exists y (B(y) land forall x (neg S(x,x) to S(y,x)))$$






                        share|cite|improve this answer











                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          There are two issues about your formula:

                          First, you use an existential quantifier for the people where there should be a univeral one, since the statement is "all people". What your formula expresses is "There is an $x$ such that if $x$ does't shave themself, the barber does." Instead, what you want to say is that "For all $x$ it holds that if $x$ doesn't shave themself, the barber does." In general, an existentially quantified implicational formula is often a sign that something is wrong.

                          Second, instead of using a constant name $b$ for the barber, it would be better to directly translate the "There is" as an existential quantifier, and you could also specify that this someone is a barber.



                          Whith this, the sentence becomes




                          There is a thing $y$ which is a barber and such that for all people $x$ who don't shave themselves, $y$ shaves $x$.




                          which translates as



                          $$exists y (B(y) land forall x (neg S(x,x) to S(y,x)))$$






                          share|cite|improve this answer











                          $endgroup$



                          There are two issues about your formula:

                          First, you use an existential quantifier for the people where there should be a univeral one, since the statement is "all people". What your formula expresses is "There is an $x$ such that if $x$ does't shave themself, the barber does." Instead, what you want to say is that "For all $x$ it holds that if $x$ doesn't shave themself, the barber does." In general, an existentially quantified implicational formula is often a sign that something is wrong.

                          Second, instead of using a constant name $b$ for the barber, it would be better to directly translate the "There is" as an existential quantifier, and you could also specify that this someone is a barber.



                          Whith this, the sentence becomes




                          There is a thing $y$ which is a barber and such that for all people $x$ who don't shave themselves, $y$ shaves $x$.




                          which translates as



                          $$exists y (B(y) land forall x (neg S(x,x) to S(y,x)))$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 1 hour ago

























                          answered 1 hour ago









                          lemontreelemontree

                          1664




                          1664






























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