How to write “There is at least 2 Cars are not the same Colour” in logic












0












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I would like to know how to write “At least 2 Cars are not the same Colour” in logic.



I know that "at least two cars" can be defined as ∃x(C(x)∧∃y(C(x)∧y≠x)). is that right? What's next then



Also, how to write "himself/herself" in logic? For example "nobody is better than himself/herself".



Appreciate for any help.






logic predicate-logic logic-translation







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    0












    $begingroup$


    I would like to know how to write “At least 2 Cars are not the same Colour” in logic.



    I know that "at least two cars" can be defined as ∃x(C(x)∧∃y(C(x)∧y≠x)). is that right? What's next then



    Also, how to write "himself/herself" in logic? For example "nobody is better than himself/herself".



    Appreciate for any help.






    logic predicate-logic logic-translation







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I would like to know how to write “At least 2 Cars are not the same Colour” in logic.



      I know that "at least two cars" can be defined as ∃x(C(x)∧∃y(C(x)∧y≠x)). is that right? What's next then



      Also, how to write "himself/herself" in logic? For example "nobody is better than himself/herself".



      Appreciate for any help.






      logic predicate-logic logic-translation







      share|cite|improve this question











      $endgroup$




      I would like to know how to write “At least 2 Cars are not the same Colour” in logic.



      I know that "at least two cars" can be defined as ∃x(C(x)∧∃y(C(x)∧y≠x)). is that right? What's next then



      Also, how to write "himself/herself" in logic? For example "nobody is better than himself/herself".



      Appreciate for any help.






      logic predicate-logic logic-translation




      logic predicate-logic logic-translation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 18:18









      Bram28

      61.6k44793




      61.6k44793










      asked Dec 10 '18 at 11:02









      BaivarasBaivaras

      6




      6






















          1 Answer
          1






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          2












          $begingroup$

          Suppose $C(x)$ denotes the colour $C$ of a car $x$. To translate "At least 2 cars are not the same colour" we could say "there are cars $x$ and $y$ that are not identical such that $C(x) ne C(y)$". In formal logic notation this is:



          $exists space x,y space left( (space x ne y) land (space C(x) ne C(y)) right)$



          or, without using $ne$, as follows:



          $exists space x,y space left( lnot (space x = y) land lnot(space C(x) = C(y)) right)$



          In fact, the qualification $x ne y$ is redundant, since $C(x) ne C(y) Rightarrow x ne y$. So we can simplify this to:



          $exists space x,y space left(lnot(C(x) = C(y)) right)$



          For the second part, if $B(x,y)$ denotes "$x$ is better than $y$" then "nobody is better than himself/herself" can be expressed as "there is no person $x$ such that $B(x,x)$". In formal logic notation this is:



          $nexists space x space B(x,x)$



          or



          $forall space x space lnot B(x,x)$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The answer is a bit misleading since you do not use boolean constants, but use colon and comma. This is at most just a suggestion how to write a correct formula, or an abbreviation of such. Yet not a formal logic notation.
            $endgroup$
            – Mad Hatter
            Dec 10 '18 at 12:18






          • 1




            $begingroup$
            @MadHatter I have changed the notation.
            $endgroup$
            – gandalf61
            Dec 10 '18 at 13:50











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Suppose $C(x)$ denotes the colour $C$ of a car $x$. To translate "At least 2 cars are not the same colour" we could say "there are cars $x$ and $y$ that are not identical such that $C(x) ne C(y)$". In formal logic notation this is:



          $exists space x,y space left( (space x ne y) land (space C(x) ne C(y)) right)$



          or, without using $ne$, as follows:



          $exists space x,y space left( lnot (space x = y) land lnot(space C(x) = C(y)) right)$



          In fact, the qualification $x ne y$ is redundant, since $C(x) ne C(y) Rightarrow x ne y$. So we can simplify this to:



          $exists space x,y space left(lnot(C(x) = C(y)) right)$



          For the second part, if $B(x,y)$ denotes "$x$ is better than $y$" then "nobody is better than himself/herself" can be expressed as "there is no person $x$ such that $B(x,x)$". In formal logic notation this is:



          $nexists space x space B(x,x)$



          or



          $forall space x space lnot B(x,x)$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The answer is a bit misleading since you do not use boolean constants, but use colon and comma. This is at most just a suggestion how to write a correct formula, or an abbreviation of such. Yet not a formal logic notation.
            $endgroup$
            – Mad Hatter
            Dec 10 '18 at 12:18






          • 1




            $begingroup$
            @MadHatter I have changed the notation.
            $endgroup$
            – gandalf61
            Dec 10 '18 at 13:50
















          2












          $begingroup$

          Suppose $C(x)$ denotes the colour $C$ of a car $x$. To translate "At least 2 cars are not the same colour" we could say "there are cars $x$ and $y$ that are not identical such that $C(x) ne C(y)$". In formal logic notation this is:



          $exists space x,y space left( (space x ne y) land (space C(x) ne C(y)) right)$



          or, without using $ne$, as follows:



          $exists space x,y space left( lnot (space x = y) land lnot(space C(x) = C(y)) right)$



          In fact, the qualification $x ne y$ is redundant, since $C(x) ne C(y) Rightarrow x ne y$. So we can simplify this to:



          $exists space x,y space left(lnot(C(x) = C(y)) right)$



          For the second part, if $B(x,y)$ denotes "$x$ is better than $y$" then "nobody is better than himself/herself" can be expressed as "there is no person $x$ such that $B(x,x)$". In formal logic notation this is:



          $nexists space x space B(x,x)$



          or



          $forall space x space lnot B(x,x)$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The answer is a bit misleading since you do not use boolean constants, but use colon and comma. This is at most just a suggestion how to write a correct formula, or an abbreviation of such. Yet not a formal logic notation.
            $endgroup$
            – Mad Hatter
            Dec 10 '18 at 12:18






          • 1




            $begingroup$
            @MadHatter I have changed the notation.
            $endgroup$
            – gandalf61
            Dec 10 '18 at 13:50














          2












          2








          2





          $begingroup$

          Suppose $C(x)$ denotes the colour $C$ of a car $x$. To translate "At least 2 cars are not the same colour" we could say "there are cars $x$ and $y$ that are not identical such that $C(x) ne C(y)$". In formal logic notation this is:



          $exists space x,y space left( (space x ne y) land (space C(x) ne C(y)) right)$



          or, without using $ne$, as follows:



          $exists space x,y space left( lnot (space x = y) land lnot(space C(x) = C(y)) right)$



          In fact, the qualification $x ne y$ is redundant, since $C(x) ne C(y) Rightarrow x ne y$. So we can simplify this to:



          $exists space x,y space left(lnot(C(x) = C(y)) right)$



          For the second part, if $B(x,y)$ denotes "$x$ is better than $y$" then "nobody is better than himself/herself" can be expressed as "there is no person $x$ such that $B(x,x)$". In formal logic notation this is:



          $nexists space x space B(x,x)$



          or



          $forall space x space lnot B(x,x)$






          share|cite|improve this answer











          $endgroup$



          Suppose $C(x)$ denotes the colour $C$ of a car $x$. To translate "At least 2 cars are not the same colour" we could say "there are cars $x$ and $y$ that are not identical such that $C(x) ne C(y)$". In formal logic notation this is:



          $exists space x,y space left( (space x ne y) land (space C(x) ne C(y)) right)$



          or, without using $ne$, as follows:



          $exists space x,y space left( lnot (space x = y) land lnot(space C(x) = C(y)) right)$



          In fact, the qualification $x ne y$ is redundant, since $C(x) ne C(y) Rightarrow x ne y$. So we can simplify this to:



          $exists space x,y space left(lnot(C(x) = C(y)) right)$



          For the second part, if $B(x,y)$ denotes "$x$ is better than $y$" then "nobody is better than himself/herself" can be expressed as "there is no person $x$ such that $B(x,x)$". In formal logic notation this is:



          $nexists space x space B(x,x)$



          or



          $forall space x space lnot B(x,x)$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 13:50

























          answered Dec 10 '18 at 11:51









          gandalf61gandalf61

          8,481725




          8,481725












          • $begingroup$
            The answer is a bit misleading since you do not use boolean constants, but use colon and comma. This is at most just a suggestion how to write a correct formula, or an abbreviation of such. Yet not a formal logic notation.
            $endgroup$
            – Mad Hatter
            Dec 10 '18 at 12:18






          • 1




            $begingroup$
            @MadHatter I have changed the notation.
            $endgroup$
            – gandalf61
            Dec 10 '18 at 13:50


















          • $begingroup$
            The answer is a bit misleading since you do not use boolean constants, but use colon and comma. This is at most just a suggestion how to write a correct formula, or an abbreviation of such. Yet not a formal logic notation.
            $endgroup$
            – Mad Hatter
            Dec 10 '18 at 12:18






          • 1




            $begingroup$
            @MadHatter I have changed the notation.
            $endgroup$
            – gandalf61
            Dec 10 '18 at 13:50
















          $begingroup$
          The answer is a bit misleading since you do not use boolean constants, but use colon and comma. This is at most just a suggestion how to write a correct formula, or an abbreviation of such. Yet not a formal logic notation.
          $endgroup$
          – Mad Hatter
          Dec 10 '18 at 12:18




          $begingroup$
          The answer is a bit misleading since you do not use boolean constants, but use colon and comma. This is at most just a suggestion how to write a correct formula, or an abbreviation of such. Yet not a formal logic notation.
          $endgroup$
          – Mad Hatter
          Dec 10 '18 at 12:18




          1




          1




          $begingroup$
          @MadHatter I have changed the notation.
          $endgroup$
          – gandalf61
          Dec 10 '18 at 13:50




          $begingroup$
          @MadHatter I have changed the notation.
          $endgroup$
          – gandalf61
          Dec 10 '18 at 13:50


















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