Negating universal and existential quantifiers.











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I'm given the formula $$neg forall x(exists yR(x,y) to R(x,f(x,x)))$$
and asked to show that it is satisfiable. I want to do this obviously by satisfying it in a structure, namely the structure $N={mathbb{N}, =, + }$, (natural numbers with addition and equality). In pushing through the first negation, I get
$$ exists xneg(exists yR(x,y) to R(x,f(x,x))).$$ That is, I want to show that there is an $x$ and there is a $y$ such that $R(x,y) to R(x,f(x,x))$ is false. Let $x$ be any number and take $y=x$, then we have $(x=y)to (x=x+x)$ is false. Does this show that the formula is satisfiable?






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  • You have to choose $xne 0,$ but that's minor. Remember the $exists y$ binds just to the $R(x,y)$, so it's not the case that you have to find an $y$ such that $R(x,y)to R(x,f(x,x))$ is false. You need to show that $exists yR(x,y)to R(x,f(x,x))$ is false. But that's fine, cause $exists y R(x,y)$ is true, and (provided you've chosen $xne 0$) $R(x,f(x,x))$ is false.
    – spaceisdarkgreen
    Dec 3 at 4:23












  • @spaceisdarkgreen Because we are using $mathbb{N}$ and $0notin mathbb{N}$, we wouldn't have to worry about $x=0$, correct?
    – Chubwagon
    Dec 3 at 4:27






  • 1




    Usually in contexts like this $0inmathbb N,$ but conventions vary.
    – spaceisdarkgreen
    Dec 3 at 5:30















up vote
0
down vote

favorite












I'm given the formula $$neg forall x(exists yR(x,y) to R(x,f(x,x)))$$
and asked to show that it is satisfiable. I want to do this obviously by satisfying it in a structure, namely the structure $N={mathbb{N}, =, + }$, (natural numbers with addition and equality). In pushing through the first negation, I get
$$ exists xneg(exists yR(x,y) to R(x,f(x,x))).$$ That is, I want to show that there is an $x$ and there is a $y$ such that $R(x,y) to R(x,f(x,x))$ is false. Let $x$ be any number and take $y=x$, then we have $(x=y)to (x=x+x)$ is false. Does this show that the formula is satisfiable?






logic







share|cite|improve this question






















  • You have to choose $xne 0,$ but that's minor. Remember the $exists y$ binds just to the $R(x,y)$, so it's not the case that you have to find an $y$ such that $R(x,y)to R(x,f(x,x))$ is false. You need to show that $exists yR(x,y)to R(x,f(x,x))$ is false. But that's fine, cause $exists y R(x,y)$ is true, and (provided you've chosen $xne 0$) $R(x,f(x,x))$ is false.
    – spaceisdarkgreen
    Dec 3 at 4:23












  • @spaceisdarkgreen Because we are using $mathbb{N}$ and $0notin mathbb{N}$, we wouldn't have to worry about $x=0$, correct?
    – Chubwagon
    Dec 3 at 4:27






  • 1




    Usually in contexts like this $0inmathbb N,$ but conventions vary.
    – spaceisdarkgreen
    Dec 3 at 5:30













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm given the formula $$neg forall x(exists yR(x,y) to R(x,f(x,x)))$$
and asked to show that it is satisfiable. I want to do this obviously by satisfying it in a structure, namely the structure $N={mathbb{N}, =, + }$, (natural numbers with addition and equality). In pushing through the first negation, I get
$$ exists xneg(exists yR(x,y) to R(x,f(x,x))).$$ That is, I want to show that there is an $x$ and there is a $y$ such that $R(x,y) to R(x,f(x,x))$ is false. Let $x$ be any number and take $y=x$, then we have $(x=y)to (x=x+x)$ is false. Does this show that the formula is satisfiable?






logic







share|cite|improve this question













I'm given the formula $$neg forall x(exists yR(x,y) to R(x,f(x,x)))$$
and asked to show that it is satisfiable. I want to do this obviously by satisfying it in a structure, namely the structure $N={mathbb{N}, =, + }$, (natural numbers with addition and equality). In pushing through the first negation, I get
$$ exists xneg(exists yR(x,y) to R(x,f(x,x))).$$ That is, I want to show that there is an $x$ and there is a $y$ such that $R(x,y) to R(x,f(x,x))$ is false. Let $x$ be any number and take $y=x$, then we have $(x=y)to (x=x+x)$ is false. Does this show that the formula is satisfiable?






logic




logic






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asked Dec 3 at 3:06









Chubwagon

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  • You have to choose $xne 0,$ but that's minor. Remember the $exists y$ binds just to the $R(x,y)$, so it's not the case that you have to find an $y$ such that $R(x,y)to R(x,f(x,x))$ is false. You need to show that $exists yR(x,y)to R(x,f(x,x))$ is false. But that's fine, cause $exists y R(x,y)$ is true, and (provided you've chosen $xne 0$) $R(x,f(x,x))$ is false.
    – spaceisdarkgreen
    Dec 3 at 4:23












  • @spaceisdarkgreen Because we are using $mathbb{N}$ and $0notin mathbb{N}$, we wouldn't have to worry about $x=0$, correct?
    – Chubwagon
    Dec 3 at 4:27






  • 1




    Usually in contexts like this $0inmathbb N,$ but conventions vary.
    – spaceisdarkgreen
    Dec 3 at 5:30


















  • You have to choose $xne 0,$ but that's minor. Remember the $exists y$ binds just to the $R(x,y)$, so it's not the case that you have to find an $y$ such that $R(x,y)to R(x,f(x,x))$ is false. You need to show that $exists yR(x,y)to R(x,f(x,x))$ is false. But that's fine, cause $exists y R(x,y)$ is true, and (provided you've chosen $xne 0$) $R(x,f(x,x))$ is false.
    – spaceisdarkgreen
    Dec 3 at 4:23












  • @spaceisdarkgreen Because we are using $mathbb{N}$ and $0notin mathbb{N}$, we wouldn't have to worry about $x=0$, correct?
    – Chubwagon
    Dec 3 at 4:27






  • 1




    Usually in contexts like this $0inmathbb N,$ but conventions vary.
    – spaceisdarkgreen
    Dec 3 at 5:30
















You have to choose $xne 0,$ but that's minor. Remember the $exists y$ binds just to the $R(x,y)$, so it's not the case that you have to find an $y$ such that $R(x,y)to R(x,f(x,x))$ is false. You need to show that $exists yR(x,y)to R(x,f(x,x))$ is false. But that's fine, cause $exists y R(x,y)$ is true, and (provided you've chosen $xne 0$) $R(x,f(x,x))$ is false.
– spaceisdarkgreen
Dec 3 at 4:23






You have to choose $xne 0,$ but that's minor. Remember the $exists y$ binds just to the $R(x,y)$, so it's not the case that you have to find an $y$ such that $R(x,y)to R(x,f(x,x))$ is false. You need to show that $exists yR(x,y)to R(x,f(x,x))$ is false. But that's fine, cause $exists y R(x,y)$ is true, and (provided you've chosen $xne 0$) $R(x,f(x,x))$ is false.
– spaceisdarkgreen
Dec 3 at 4:23














@spaceisdarkgreen Because we are using $mathbb{N}$ and $0notin mathbb{N}$, we wouldn't have to worry about $x=0$, correct?
– Chubwagon
Dec 3 at 4:27




@spaceisdarkgreen Because we are using $mathbb{N}$ and $0notin mathbb{N}$, we wouldn't have to worry about $x=0$, correct?
– Chubwagon
Dec 3 at 4:27




1




1




Usually in contexts like this $0inmathbb N,$ but conventions vary.
– spaceisdarkgreen
Dec 3 at 5:30




Usually in contexts like this $0inmathbb N,$ but conventions vary.
– spaceisdarkgreen
Dec 3 at 5:30










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By operator precedence conventions (i.e. order of operations), conditional operation takes precedence over quantifier binding. That is the existential is the antecedant of the conditional.   So the quivalent statements you want are:



$begin{split}&neg forall x~(color{silver}(exists y~R(x,y)color{silver}) to R(x,f(x,x))) \equiv\& exists x~lnot((exists y~R(x,y))to R(x,f(x,x)))\equiv\& exists x~((exists y~R(x,y))landlnot R(x,f(x,x)))end{split}$



Thus you need to show that there can be some natural number, $x$, where $R(x,f(x,x))$ is false, but there is also some natural number, $y$, where $R(x,y)$ is true.



Which will depend on what are the bivariate function $f$ and relation $R$. If you use $+$ and $=$: $$exists x~((exists y~x=y)land(xneq x+x))$$






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    By operator precedence conventions (i.e. order of operations), conditional operation takes precedence over quantifier binding. That is the existential is the antecedant of the conditional.   So the quivalent statements you want are:



    $begin{split}&neg forall x~(color{silver}(exists y~R(x,y)color{silver}) to R(x,f(x,x))) \equiv\& exists x~lnot((exists y~R(x,y))to R(x,f(x,x)))\equiv\& exists x~((exists y~R(x,y))landlnot R(x,f(x,x)))end{split}$



    Thus you need to show that there can be some natural number, $x$, where $R(x,f(x,x))$ is false, but there is also some natural number, $y$, where $R(x,y)$ is true.



    Which will depend on what are the bivariate function $f$ and relation $R$. If you use $+$ and $=$: $$exists x~((exists y~x=y)land(xneq x+x))$$






    share|cite|improve this answer

























      up vote
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      down vote













      By operator precedence conventions (i.e. order of operations), conditional operation takes precedence over quantifier binding. That is the existential is the antecedant of the conditional.   So the quivalent statements you want are:



      $begin{split}&neg forall x~(color{silver}(exists y~R(x,y)color{silver}) to R(x,f(x,x))) \equiv\& exists x~lnot((exists y~R(x,y))to R(x,f(x,x)))\equiv\& exists x~((exists y~R(x,y))landlnot R(x,f(x,x)))end{split}$



      Thus you need to show that there can be some natural number, $x$, where $R(x,f(x,x))$ is false, but there is also some natural number, $y$, where $R(x,y)$ is true.



      Which will depend on what are the bivariate function $f$ and relation $R$. If you use $+$ and $=$: $$exists x~((exists y~x=y)land(xneq x+x))$$






      share|cite|improve this answer























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        down vote










        up vote
        0
        down vote









        By operator precedence conventions (i.e. order of operations), conditional operation takes precedence over quantifier binding. That is the existential is the antecedant of the conditional.   So the quivalent statements you want are:



        $begin{split}&neg forall x~(color{silver}(exists y~R(x,y)color{silver}) to R(x,f(x,x))) \equiv\& exists x~lnot((exists y~R(x,y))to R(x,f(x,x)))\equiv\& exists x~((exists y~R(x,y))landlnot R(x,f(x,x)))end{split}$



        Thus you need to show that there can be some natural number, $x$, where $R(x,f(x,x))$ is false, but there is also some natural number, $y$, where $R(x,y)$ is true.



        Which will depend on what are the bivariate function $f$ and relation $R$. If you use $+$ and $=$: $$exists x~((exists y~x=y)land(xneq x+x))$$






        share|cite|improve this answer












        By operator precedence conventions (i.e. order of operations), conditional operation takes precedence over quantifier binding. That is the existential is the antecedant of the conditional.   So the quivalent statements you want are:



        $begin{split}&neg forall x~(color{silver}(exists y~R(x,y)color{silver}) to R(x,f(x,x))) \equiv\& exists x~lnot((exists y~R(x,y))to R(x,f(x,x)))\equiv\& exists x~((exists y~R(x,y))landlnot R(x,f(x,x)))end{split}$



        Thus you need to show that there can be some natural number, $x$, where $R(x,f(x,x))$ is false, but there is also some natural number, $y$, where $R(x,y)$ is true.



        Which will depend on what are the bivariate function $f$ and relation $R$. If you use $+$ and $=$: $$exists x~((exists y~x=y)land(xneq x+x))$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 at 23:06









        Graham Kemp

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