Formal definition of truth function












0












$begingroup$


As Mendelson say in Introduction to Mathematical Logic:




As we have seen, any statement form containing $n$ distinct statement letters determines a corresponding truth function of $n$ arguments.




Thus, it must exist a function $mathcal{F}: {text{statement forms}} to {text{truth function}}$, where the set of statement forms is given by the definition in page 3 and the set of true function is
$$bigcup_{ninmathbb{N}} {T,F}^{{T,F}^n} = bigcup_{ninmathbb{N}} 2^{2^n}.$$



I have an idea of how to define this function, but I want to know if there exists a book or an article where I can find a formal and precise definition of that function?










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  • $begingroup$
    See similar post: the-number-of-n-variable-preposition-function.
    $endgroup$
    – Mauro ALLEGRANZA
    Oct 30 '17 at 16:14
















0












$begingroup$


As Mendelson say in Introduction to Mathematical Logic:




As we have seen, any statement form containing $n$ distinct statement letters determines a corresponding truth function of $n$ arguments.




Thus, it must exist a function $mathcal{F}: {text{statement forms}} to {text{truth function}}$, where the set of statement forms is given by the definition in page 3 and the set of true function is
$$bigcup_{ninmathbb{N}} {T,F}^{{T,F}^n} = bigcup_{ninmathbb{N}} 2^{2^n}.$$



I have an idea of how to define this function, but I want to know if there exists a book or an article where I can find a formal and precise definition of that function?










share|cite|improve this question









$endgroup$












  • $begingroup$
    See similar post: the-number-of-n-variable-preposition-function.
    $endgroup$
    – Mauro ALLEGRANZA
    Oct 30 '17 at 16:14














0












0








0





$begingroup$


As Mendelson say in Introduction to Mathematical Logic:




As we have seen, any statement form containing $n$ distinct statement letters determines a corresponding truth function of $n$ arguments.




Thus, it must exist a function $mathcal{F}: {text{statement forms}} to {text{truth function}}$, where the set of statement forms is given by the definition in page 3 and the set of true function is
$$bigcup_{ninmathbb{N}} {T,F}^{{T,F}^n} = bigcup_{ninmathbb{N}} 2^{2^n}.$$



I have an idea of how to define this function, but I want to know if there exists a book or an article where I can find a formal and precise definition of that function?










share|cite|improve this question









$endgroup$




As Mendelson say in Introduction to Mathematical Logic:




As we have seen, any statement form containing $n$ distinct statement letters determines a corresponding truth function of $n$ arguments.




Thus, it must exist a function $mathcal{F}: {text{statement forms}} to {text{truth function}}$, where the set of statement forms is given by the definition in page 3 and the set of true function is
$$bigcup_{ninmathbb{N}} {T,F}^{{T,F}^n} = bigcup_{ninmathbb{N}} 2^{2^n}.$$



I have an idea of how to define this function, but I want to know if there exists a book or an article where I can find a formal and precise definition of that function?







logic propositional-calculus






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asked Oct 30 '17 at 15:31









andrebantandrebant

12611




12611












  • $begingroup$
    See similar post: the-number-of-n-variable-preposition-function.
    $endgroup$
    – Mauro ALLEGRANZA
    Oct 30 '17 at 16:14


















  • $begingroup$
    See similar post: the-number-of-n-variable-preposition-function.
    $endgroup$
    – Mauro ALLEGRANZA
    Oct 30 '17 at 16:14
















$begingroup$
See similar post: the-number-of-n-variable-preposition-function.
$endgroup$
– Mauro ALLEGRANZA
Oct 30 '17 at 16:14




$begingroup$
See similar post: the-number-of-n-variable-preposition-function.
$endgroup$
– Mauro ALLEGRANZA
Oct 30 '17 at 16:14










2 Answers
2






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oldest

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1












$begingroup$

But Mendelson's discussion is already entirely precise. He says (2nd edition, but I think this is preserved through later editions):




For every assignment of truth values T or F to the statement letters
occurring in a statement form, there corresponds, by virtue of the
truth tables for the propositional connectives, a truth value for the
statement form. Thus, each statement form determines a truth function.




What could be clearer or more precise than that? Going "formal" wouldn't make it more precise -- it would merely translate Mendelson's crisp description of the map between statement forms and corresponding truth-functions into a much less transparent symbolic description of the same map.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    This is my own idea:



    Notation:





    • $S$ - the set of statement letters


    • $W$ the set of statement forms


    • $mathrm{Sub}(A)$ - the set of all subformulas of $Ain W$.


    It is a known fact that each function $nu_0:Srightarrow{T,F}$ can be uniquely extended to a function $nu:Wrightarrow{T,F}$ in such way that the vaules on negation, conjuction etc. depend on values on constituents.



    Another thing we need is that for any $Ain W$, the set $mathrm{Sub}(A)cap S$ is nonempty and finite. This is proved by structural induction.



    From the definition, $S$ is countable. Let $S={a_1,a_2,a_3,ldots}$, where $(a_n)$ is injective. We define as well an order on $S$ by condition $a_ile a_j :iff ile j$



    Now let's go to defining $mathcal{F}$. Fix $Ain W$. The set $mathrm{Sub}(A)cap S$ is a finite,nonempty subset of $S$. Therefore it can be enumerated: $mathrm{Sub}(A)cap S={b_1,ldots,b_n}$ for some $ninmathbb{N}$ and injective $(b_k)_{kle n}$ (to avoid axiom of choice $(b_k)_{kle n}$ can be uniquely chosen to be monotonic). We will define $xi:{T,F}^nrightarrow{T,F}$ and then set $mathcal{F}(A):=xi$. Fix $xin{T,F}^n$ and set $nu_0:Srightarrow{T,F}$ by conditions: $nu_0(a):=F$ for $ain Ssetminus{b_1,ldots,b_n}$ and if $ain{b_1,ldots,b_n}$ then $a=b_i$ for some $i$ and we set $nu_0(a):=x_i$. $v_0$ can be uniquely extended to $nu:Wrightarrow{T,F}$ and we set $xi(x):=nu(A)$.



    An important thing to note is that $mathcal{F}$ depends on the choice of enumeration $(a_n)$.






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      But Mendelson's discussion is already entirely precise. He says (2nd edition, but I think this is preserved through later editions):




      For every assignment of truth values T or F to the statement letters
      occurring in a statement form, there corresponds, by virtue of the
      truth tables for the propositional connectives, a truth value for the
      statement form. Thus, each statement form determines a truth function.




      What could be clearer or more precise than that? Going "formal" wouldn't make it more precise -- it would merely translate Mendelson's crisp description of the map between statement forms and corresponding truth-functions into a much less transparent symbolic description of the same map.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        But Mendelson's discussion is already entirely precise. He says (2nd edition, but I think this is preserved through later editions):




        For every assignment of truth values T or F to the statement letters
        occurring in a statement form, there corresponds, by virtue of the
        truth tables for the propositional connectives, a truth value for the
        statement form. Thus, each statement form determines a truth function.




        What could be clearer or more precise than that? Going "formal" wouldn't make it more precise -- it would merely translate Mendelson's crisp description of the map between statement forms and corresponding truth-functions into a much less transparent symbolic description of the same map.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          But Mendelson's discussion is already entirely precise. He says (2nd edition, but I think this is preserved through later editions):




          For every assignment of truth values T or F to the statement letters
          occurring in a statement form, there corresponds, by virtue of the
          truth tables for the propositional connectives, a truth value for the
          statement form. Thus, each statement form determines a truth function.




          What could be clearer or more precise than that? Going "formal" wouldn't make it more precise -- it would merely translate Mendelson's crisp description of the map between statement forms and corresponding truth-functions into a much less transparent symbolic description of the same map.






          share|cite|improve this answer









          $endgroup$



          But Mendelson's discussion is already entirely precise. He says (2nd edition, but I think this is preserved through later editions):




          For every assignment of truth values T or F to the statement letters
          occurring in a statement form, there corresponds, by virtue of the
          truth tables for the propositional connectives, a truth value for the
          statement form. Thus, each statement form determines a truth function.




          What could be clearer or more precise than that? Going "formal" wouldn't make it more precise -- it would merely translate Mendelson's crisp description of the map between statement forms and corresponding truth-functions into a much less transparent symbolic description of the same map.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 31 '17 at 22:33









          Peter SmithPeter Smith

          40.9k340120




          40.9k340120























              0












              $begingroup$

              This is my own idea:



              Notation:





              • $S$ - the set of statement letters


              • $W$ the set of statement forms


              • $mathrm{Sub}(A)$ - the set of all subformulas of $Ain W$.


              It is a known fact that each function $nu_0:Srightarrow{T,F}$ can be uniquely extended to a function $nu:Wrightarrow{T,F}$ in such way that the vaules on negation, conjuction etc. depend on values on constituents.



              Another thing we need is that for any $Ain W$, the set $mathrm{Sub}(A)cap S$ is nonempty and finite. This is proved by structural induction.



              From the definition, $S$ is countable. Let $S={a_1,a_2,a_3,ldots}$, where $(a_n)$ is injective. We define as well an order on $S$ by condition $a_ile a_j :iff ile j$



              Now let's go to defining $mathcal{F}$. Fix $Ain W$. The set $mathrm{Sub}(A)cap S$ is a finite,nonempty subset of $S$. Therefore it can be enumerated: $mathrm{Sub}(A)cap S={b_1,ldots,b_n}$ for some $ninmathbb{N}$ and injective $(b_k)_{kle n}$ (to avoid axiom of choice $(b_k)_{kle n}$ can be uniquely chosen to be monotonic). We will define $xi:{T,F}^nrightarrow{T,F}$ and then set $mathcal{F}(A):=xi$. Fix $xin{T,F}^n$ and set $nu_0:Srightarrow{T,F}$ by conditions: $nu_0(a):=F$ for $ain Ssetminus{b_1,ldots,b_n}$ and if $ain{b_1,ldots,b_n}$ then $a=b_i$ for some $i$ and we set $nu_0(a):=x_i$. $v_0$ can be uniquely extended to $nu:Wrightarrow{T,F}$ and we set $xi(x):=nu(A)$.



              An important thing to note is that $mathcal{F}$ depends on the choice of enumeration $(a_n)$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                This is my own idea:



                Notation:





                • $S$ - the set of statement letters


                • $W$ the set of statement forms


                • $mathrm{Sub}(A)$ - the set of all subformulas of $Ain W$.


                It is a known fact that each function $nu_0:Srightarrow{T,F}$ can be uniquely extended to a function $nu:Wrightarrow{T,F}$ in such way that the vaules on negation, conjuction etc. depend on values on constituents.



                Another thing we need is that for any $Ain W$, the set $mathrm{Sub}(A)cap S$ is nonempty and finite. This is proved by structural induction.



                From the definition, $S$ is countable. Let $S={a_1,a_2,a_3,ldots}$, where $(a_n)$ is injective. We define as well an order on $S$ by condition $a_ile a_j :iff ile j$



                Now let's go to defining $mathcal{F}$. Fix $Ain W$. The set $mathrm{Sub}(A)cap S$ is a finite,nonempty subset of $S$. Therefore it can be enumerated: $mathrm{Sub}(A)cap S={b_1,ldots,b_n}$ for some $ninmathbb{N}$ and injective $(b_k)_{kle n}$ (to avoid axiom of choice $(b_k)_{kle n}$ can be uniquely chosen to be monotonic). We will define $xi:{T,F}^nrightarrow{T,F}$ and then set $mathcal{F}(A):=xi$. Fix $xin{T,F}^n$ and set $nu_0:Srightarrow{T,F}$ by conditions: $nu_0(a):=F$ for $ain Ssetminus{b_1,ldots,b_n}$ and if $ain{b_1,ldots,b_n}$ then $a=b_i$ for some $i$ and we set $nu_0(a):=x_i$. $v_0$ can be uniquely extended to $nu:Wrightarrow{T,F}$ and we set $xi(x):=nu(A)$.



                An important thing to note is that $mathcal{F}$ depends on the choice of enumeration $(a_n)$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  This is my own idea:



                  Notation:





                  • $S$ - the set of statement letters


                  • $W$ the set of statement forms


                  • $mathrm{Sub}(A)$ - the set of all subformulas of $Ain W$.


                  It is a known fact that each function $nu_0:Srightarrow{T,F}$ can be uniquely extended to a function $nu:Wrightarrow{T,F}$ in such way that the vaules on negation, conjuction etc. depend on values on constituents.



                  Another thing we need is that for any $Ain W$, the set $mathrm{Sub}(A)cap S$ is nonempty and finite. This is proved by structural induction.



                  From the definition, $S$ is countable. Let $S={a_1,a_2,a_3,ldots}$, where $(a_n)$ is injective. We define as well an order on $S$ by condition $a_ile a_j :iff ile j$



                  Now let's go to defining $mathcal{F}$. Fix $Ain W$. The set $mathrm{Sub}(A)cap S$ is a finite,nonempty subset of $S$. Therefore it can be enumerated: $mathrm{Sub}(A)cap S={b_1,ldots,b_n}$ for some $ninmathbb{N}$ and injective $(b_k)_{kle n}$ (to avoid axiom of choice $(b_k)_{kle n}$ can be uniquely chosen to be monotonic). We will define $xi:{T,F}^nrightarrow{T,F}$ and then set $mathcal{F}(A):=xi$. Fix $xin{T,F}^n$ and set $nu_0:Srightarrow{T,F}$ by conditions: $nu_0(a):=F$ for $ain Ssetminus{b_1,ldots,b_n}$ and if $ain{b_1,ldots,b_n}$ then $a=b_i$ for some $i$ and we set $nu_0(a):=x_i$. $v_0$ can be uniquely extended to $nu:Wrightarrow{T,F}$ and we set $xi(x):=nu(A)$.



                  An important thing to note is that $mathcal{F}$ depends on the choice of enumeration $(a_n)$.






                  share|cite|improve this answer











                  $endgroup$



                  This is my own idea:



                  Notation:





                  • $S$ - the set of statement letters


                  • $W$ the set of statement forms


                  • $mathrm{Sub}(A)$ - the set of all subformulas of $Ain W$.


                  It is a known fact that each function $nu_0:Srightarrow{T,F}$ can be uniquely extended to a function $nu:Wrightarrow{T,F}$ in such way that the vaules on negation, conjuction etc. depend on values on constituents.



                  Another thing we need is that for any $Ain W$, the set $mathrm{Sub}(A)cap S$ is nonempty and finite. This is proved by structural induction.



                  From the definition, $S$ is countable. Let $S={a_1,a_2,a_3,ldots}$, where $(a_n)$ is injective. We define as well an order on $S$ by condition $a_ile a_j :iff ile j$



                  Now let's go to defining $mathcal{F}$. Fix $Ain W$. The set $mathrm{Sub}(A)cap S$ is a finite,nonempty subset of $S$. Therefore it can be enumerated: $mathrm{Sub}(A)cap S={b_1,ldots,b_n}$ for some $ninmathbb{N}$ and injective $(b_k)_{kle n}$ (to avoid axiom of choice $(b_k)_{kle n}$ can be uniquely chosen to be monotonic). We will define $xi:{T,F}^nrightarrow{T,F}$ and then set $mathcal{F}(A):=xi$. Fix $xin{T,F}^n$ and set $nu_0:Srightarrow{T,F}$ by conditions: $nu_0(a):=F$ for $ain Ssetminus{b_1,ldots,b_n}$ and if $ain{b_1,ldots,b_n}$ then $a=b_i$ for some $i$ and we set $nu_0(a):=x_i$. $v_0$ can be uniquely extended to $nu:Wrightarrow{T,F}$ and we set $xi(x):=nu(A)$.



                  An important thing to note is that $mathcal{F}$ depends on the choice of enumeration $(a_n)$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 23 at 19:06

























                  answered Jan 23 at 18:45









                  KulistyKulisty

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                  419216






























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