How to define substitution using ZFC












0












$begingroup$


One question I've had regarding ZFC is how to define substitution. I cannot see how it's possible, despite the frequent use of substitution within both pure and applied mathematics.



Just to be clear, I define substitution as follows: for all well-defined mathematical objects $a$ and $b$ that are equal, and any function $f$ such that $f(a)$ and $f(b)$ are well-defined, $f(a) = f(b)$.
$$ forall a forall b forall f [(a = b , wedge f(a),f(b) text{ are defined}) Rightarrow f(a) = f(b)]$$



Any ideas?






substitution foundations







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$endgroup$

















    0












    $begingroup$


    One question I've had regarding ZFC is how to define substitution. I cannot see how it's possible, despite the frequent use of substitution within both pure and applied mathematics.



    Just to be clear, I define substitution as follows: for all well-defined mathematical objects $a$ and $b$ that are equal, and any function $f$ such that $f(a)$ and $f(b)$ are well-defined, $f(a) = f(b)$.
    $$ forall a forall b forall f [(a = b , wedge f(a),f(b) text{ are defined}) Rightarrow f(a) = f(b)]$$



    Any ideas?






    substitution foundations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      One question I've had regarding ZFC is how to define substitution. I cannot see how it's possible, despite the frequent use of substitution within both pure and applied mathematics.



      Just to be clear, I define substitution as follows: for all well-defined mathematical objects $a$ and $b$ that are equal, and any function $f$ such that $f(a)$ and $f(b)$ are well-defined, $f(a) = f(b)$.
      $$ forall a forall b forall f [(a = b , wedge f(a),f(b) text{ are defined}) Rightarrow f(a) = f(b)]$$



      Any ideas?






      substitution foundations







      share|cite|improve this question









      $endgroup$




      One question I've had regarding ZFC is how to define substitution. I cannot see how it's possible, despite the frequent use of substitution within both pure and applied mathematics.



      Just to be clear, I define substitution as follows: for all well-defined mathematical objects $a$ and $b$ that are equal, and any function $f$ such that $f(a)$ and $f(b)$ are well-defined, $f(a) = f(b)$.
      $$ forall a forall b forall f [(a = b , wedge f(a),f(b) text{ are defined}) Rightarrow f(a) = f(b)]$$



      Any ideas?






      substitution foundations




      substitution foundations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 31 '18 at 12:22









      HarrisonOHarrisonO

      707




      707






















          2 Answers
          2






          active

          oldest

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          3












          $begingroup$

          Substitution is "governed" by the axioms for equality.



          We can substitute equals into a formula $varphi$ :




          $a = b → (φ[a/x] → φ[b/x])$, for any formula $φ(x)$,




          or we can substitute into a function $f$ :




          $a=b to f(ldots,a,ldots) = f(ldots,b,ldots)$, for any function symbol $f$.




          Example in $mathsf {ZF}$, using the formula $(z in x)$ as $φ(x)$ :





          $a = b to ((z in a) to (z in b))$.








          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So is it not possible to define substitution using the ZF or ZFC axioms, requiring the axioms for equality? Otherwise, can you please tell me the exact axioms needed. Thanks.
            $endgroup$
            – HarrisonO
            Dec 31 '18 at 23:50






          • 1




            $begingroup$
            @HarrisonO - $mathsf {ZF}$ and $mathsf {ZFC}$ are "axiom systems formulated in first-order logic with equality and with only one binary relation symbol $∈$ for membership
            $endgroup$
            – Mauro ALLEGRANZA
            Jan 2 at 10:16





















          0












          $begingroup$

          You can actually formulate ZFC in FOL without equality: with extensionality reading



          $ forall z ( z in x iff z in y ) implies forall z ( x in z iff y in z ) $



          define a binary predicate to mean the consequent in the last sentence



          $ xRy :iff forall z ( x in z iff y in z ) $



          reflexivity, transitivity and symmetry following easily from FOL alone; and then with the usual comprehension/separation scheme and weak-pair



          $ forall x exists yforall z(z in y iff z in x wedge varphi) $



          $ forall x forall y exists z (x in z wedge y in z) $



          prove substitution for $ R $, ie:



          $ xRy implies (varphi x iff varphi y) $






          share|cite|improve this answer









          $endgroup$














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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Substitution is "governed" by the axioms for equality.



            We can substitute equals into a formula $varphi$ :




            $a = b → (φ[a/x] → φ[b/x])$, for any formula $φ(x)$,




            or we can substitute into a function $f$ :




            $a=b to f(ldots,a,ldots) = f(ldots,b,ldots)$, for any function symbol $f$.




            Example in $mathsf {ZF}$, using the formula $(z in x)$ as $φ(x)$ :





            $a = b to ((z in a) to (z in b))$.








            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So is it not possible to define substitution using the ZF or ZFC axioms, requiring the axioms for equality? Otherwise, can you please tell me the exact axioms needed. Thanks.
              $endgroup$
              – HarrisonO
              Dec 31 '18 at 23:50






            • 1




              $begingroup$
              @HarrisonO - $mathsf {ZF}$ and $mathsf {ZFC}$ are "axiom systems formulated in first-order logic with equality and with only one binary relation symbol $∈$ for membership
              $endgroup$
              – Mauro ALLEGRANZA
              Jan 2 at 10:16


















            3












            $begingroup$

            Substitution is "governed" by the axioms for equality.



            We can substitute equals into a formula $varphi$ :




            $a = b → (φ[a/x] → φ[b/x])$, for any formula $φ(x)$,




            or we can substitute into a function $f$ :




            $a=b to f(ldots,a,ldots) = f(ldots,b,ldots)$, for any function symbol $f$.




            Example in $mathsf {ZF}$, using the formula $(z in x)$ as $φ(x)$ :





            $a = b to ((z in a) to (z in b))$.








            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So is it not possible to define substitution using the ZF or ZFC axioms, requiring the axioms for equality? Otherwise, can you please tell me the exact axioms needed. Thanks.
              $endgroup$
              – HarrisonO
              Dec 31 '18 at 23:50






            • 1




              $begingroup$
              @HarrisonO - $mathsf {ZF}$ and $mathsf {ZFC}$ are "axiom systems formulated in first-order logic with equality and with only one binary relation symbol $∈$ for membership
              $endgroup$
              – Mauro ALLEGRANZA
              Jan 2 at 10:16
















            3












            3








            3





            $begingroup$

            Substitution is "governed" by the axioms for equality.



            We can substitute equals into a formula $varphi$ :




            $a = b → (φ[a/x] → φ[b/x])$, for any formula $φ(x)$,




            or we can substitute into a function $f$ :




            $a=b to f(ldots,a,ldots) = f(ldots,b,ldots)$, for any function symbol $f$.




            Example in $mathsf {ZF}$, using the formula $(z in x)$ as $φ(x)$ :





            $a = b to ((z in a) to (z in b))$.








            share|cite|improve this answer









            $endgroup$



            Substitution is "governed" by the axioms for equality.



            We can substitute equals into a formula $varphi$ :




            $a = b → (φ[a/x] → φ[b/x])$, for any formula $φ(x)$,




            or we can substitute into a function $f$ :




            $a=b to f(ldots,a,ldots) = f(ldots,b,ldots)$, for any function symbol $f$.




            Example in $mathsf {ZF}$, using the formula $(z in x)$ as $φ(x)$ :





            $a = b to ((z in a) to (z in b))$.









            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 31 '18 at 12:42









            Mauro ALLEGRANZAMauro ALLEGRANZA

            68.3k449118




            68.3k449118












            • $begingroup$
              So is it not possible to define substitution using the ZF or ZFC axioms, requiring the axioms for equality? Otherwise, can you please tell me the exact axioms needed. Thanks.
              $endgroup$
              – HarrisonO
              Dec 31 '18 at 23:50






            • 1




              $begingroup$
              @HarrisonO - $mathsf {ZF}$ and $mathsf {ZFC}$ are "axiom systems formulated in first-order logic with equality and with only one binary relation symbol $∈$ for membership
              $endgroup$
              – Mauro ALLEGRANZA
              Jan 2 at 10:16




















            • $begingroup$
              So is it not possible to define substitution using the ZF or ZFC axioms, requiring the axioms for equality? Otherwise, can you please tell me the exact axioms needed. Thanks.
              $endgroup$
              – HarrisonO
              Dec 31 '18 at 23:50






            • 1




              $begingroup$
              @HarrisonO - $mathsf {ZF}$ and $mathsf {ZFC}$ are "axiom systems formulated in first-order logic with equality and with only one binary relation symbol $∈$ for membership
              $endgroup$
              – Mauro ALLEGRANZA
              Jan 2 at 10:16


















            $begingroup$
            So is it not possible to define substitution using the ZF or ZFC axioms, requiring the axioms for equality? Otherwise, can you please tell me the exact axioms needed. Thanks.
            $endgroup$
            – HarrisonO
            Dec 31 '18 at 23:50




            $begingroup$
            So is it not possible to define substitution using the ZF or ZFC axioms, requiring the axioms for equality? Otherwise, can you please tell me the exact axioms needed. Thanks.
            $endgroup$
            – HarrisonO
            Dec 31 '18 at 23:50




            1




            1




            $begingroup$
            @HarrisonO - $mathsf {ZF}$ and $mathsf {ZFC}$ are "axiom systems formulated in first-order logic with equality and with only one binary relation symbol $∈$ for membership
            $endgroup$
            – Mauro ALLEGRANZA
            Jan 2 at 10:16






            $begingroup$
            @HarrisonO - $mathsf {ZF}$ and $mathsf {ZFC}$ are "axiom systems formulated in first-order logic with equality and with only one binary relation symbol $∈$ for membership
            $endgroup$
            – Mauro ALLEGRANZA
            Jan 2 at 10:16













            0












            $begingroup$

            You can actually formulate ZFC in FOL without equality: with extensionality reading



            $ forall z ( z in x iff z in y ) implies forall z ( x in z iff y in z ) $



            define a binary predicate to mean the consequent in the last sentence



            $ xRy :iff forall z ( x in z iff y in z ) $



            reflexivity, transitivity and symmetry following easily from FOL alone; and then with the usual comprehension/separation scheme and weak-pair



            $ forall x exists yforall z(z in y iff z in x wedge varphi) $



            $ forall x forall y exists z (x in z wedge y in z) $



            prove substitution for $ R $, ie:



            $ xRy implies (varphi x iff varphi y) $






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              You can actually formulate ZFC in FOL without equality: with extensionality reading



              $ forall z ( z in x iff z in y ) implies forall z ( x in z iff y in z ) $



              define a binary predicate to mean the consequent in the last sentence



              $ xRy :iff forall z ( x in z iff y in z ) $



              reflexivity, transitivity and symmetry following easily from FOL alone; and then with the usual comprehension/separation scheme and weak-pair



              $ forall x exists yforall z(z in y iff z in x wedge varphi) $



              $ forall x forall y exists z (x in z wedge y in z) $



              prove substitution for $ R $, ie:



              $ xRy implies (varphi x iff varphi y) $






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                You can actually formulate ZFC in FOL without equality: with extensionality reading



                $ forall z ( z in x iff z in y ) implies forall z ( x in z iff y in z ) $



                define a binary predicate to mean the consequent in the last sentence



                $ xRy :iff forall z ( x in z iff y in z ) $



                reflexivity, transitivity and symmetry following easily from FOL alone; and then with the usual comprehension/separation scheme and weak-pair



                $ forall x exists yforall z(z in y iff z in x wedge varphi) $



                $ forall x forall y exists z (x in z wedge y in z) $



                prove substitution for $ R $, ie:



                $ xRy implies (varphi x iff varphi y) $






                share|cite|improve this answer









                $endgroup$



                You can actually formulate ZFC in FOL without equality: with extensionality reading



                $ forall z ( z in x iff z in y ) implies forall z ( x in z iff y in z ) $



                define a binary predicate to mean the consequent in the last sentence



                $ xRy :iff forall z ( x in z iff y in z ) $



                reflexivity, transitivity and symmetry following easily from FOL alone; and then with the usual comprehension/separation scheme and weak-pair



                $ forall x exists yforall z(z in y iff z in x wedge varphi) $



                $ forall x forall y exists z (x in z wedge y in z) $



                prove substitution for $ R $, ie:



                $ xRy implies (varphi x iff varphi y) $







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 15 at 1:38









                a.c.brunoa.c.bruno

                48639




                48639






























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