THEOREM

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   ALL(x):ALL(y):ALL(fs):[Set(x) & Set(y) & Set(fs)

   & EXIST(a):a e x

   & ALL(f):[f e fs <=> ALL(a):[a e x => f(a) e y]]

   & ~EXIST(a):a e y

   => ~EXIST(f):f e fs]

 

By Dan Christensen

2022-01-16

 

This machine-verified, formal proof was written with the aid of the author's DC Proof 2.0 software available at http://www.dcproof.com

 

 

PROOF

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Proposed Function Space axiom (functions of one variable)

 

1     ALL(a):ALL(b):[Set(a) & EXIST(c):c e a & Set(b) => EXIST(c):[Set(c) & ALL(f):[f e c <=> ALL(d):[d e a => f(d) e b]]]]

      Axiom

 

   

    Suppose...

 

      2     Set(x) & Set(y) & Set(fs)

         & EXIST(a):a e x

         & ALL(f):[f e fs <=> ALL(a):[a e x => f(a) e y]]

         & ~EXIST(a):a e y

            Premise

 

      3     Set(x)

            Split, 2

 

      4     Set(y)

            Split, 2

 

      5     Set(fs)

            Split, 2

 

      6     EXIST(a):a e x

            Split, 2

 

      7     ALL(f):[f e fs <=> ALL(a):[a e x => f(a) e y]]

            Split, 2

 

      8     ~EXIST(a):a e y

            Split, 2

 

      9     z e x

            E Spec, 6

 

      10   ~~ALL(a):~a e y

            Quant, 8

 

      11   ALL(a):~a e y

            Rem DNeg, 10

 

        

         Prove: ~EXIST(f):f e fs

        

         Suppose to the contrary...

 

            12   f e fs

                  Premise

 

            13   f e fs <=> ALL(a):[a e x => f(a) e y]

                  U Spec, 7

 

            14   [f e fs => ALL(a):[a e x => f(a) e y]]

             & [ALL(a):[a e x => f(a) e y] => f e fs]

                  Iff-And, 13

 

            15   f e fs => ALL(a):[a e x => f(a) e y]

                  Split, 14

 

            16   ALL(a):[a e x => f(a) e y]

                  Detach, 15, 12

 

            17   z e x => f(z) e y

                  U Spec, 16

 

            18   f(z) e y

                  Detach, 17, 9

 

            19   ~f(z) e y

                  U Spec, 11, 18

 

            20   f(z) e y & ~f(z) e y

                  Join, 18, 19

 

    As Required:

 

      21   ~EXIST(f):f e fs

            4 Conclusion, 12

 

As Required:

 

22   ALL(x):ALL(y):ALL(fs):[Set(x) & Set(y) & Set(fs)

    & EXIST(a):a e x

    & ALL(f):[f e fs <=> ALL(a):[a e x => f(a) e y]]

    & ~EXIST(a):a e y

    => ~EXIST(f):f e fs]

      4 Conclusion, 2