THEOREM

*******

 

 

De Morgan's Law:  A | B <=> ~[~A & ~B]

 

The proof uses only the following rules of inference that give the essential properties of the OR-operator ('|'):

 

1. Or Not:  A | ~A  (Law of the Excluded Middle)  (line 20)

 

2. Arb Or:  A  --->  A | B  or  B | A   (lines 22, 30)

 

3. Cases:  P1 | P2 | P3 | ... Pn  and P1 => Q, P2 => Q, P3 => Q, ... Pn => Q  --->  Q  (lines 17, 33)

 

 

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

 

Dan Christensen

2022-06-03

 

 

PROOF

*****

 

    '=>'

   

    Prove: A | B => ~[~A & ~B]

   

    Suppose (two cases to consider)...

 

      1     A | B

            Premise

 

         Case 1

        

         Prove: A => ~[~A & ~B]

        

         Suppose...

 

            2     A

                  Premise

 

             Prove: ~[~A & ~B]

            

             Suppose to the contrary...

 

                  3     ~A & ~B

                        Premise

 

                  4     ~A

                        Split, 3

 

                  5     ~B

                        Split, 3

 

                  6     A & ~A

                        Join, 2, 4

 

         As Required:

 

            7     ~[~A & ~B]

                  4 Conclusion, 3

 

    Case 1

   

    As Required:

 

      8     A => ~[~A & ~B]

            4 Conclusion, 2

 

         Case 2

        

         Prove: B => ~[~A & ~B]

        

         Suppose...

 

            9     B

                  Premise

 

             Prove: ~[~A & ~B]

            

             Suppose to the contrary...

 

                  10   ~A & ~B

                        Premise

 

                  11   ~A

                        Split, 10

 

                  12   ~B

                        Split, 10

 

                  13   B & ~B

                        Join, 9, 12

 

         As Required:

 

            14   ~[~A & ~B]

                  4 Conclusion, 10

 

    Case 2

   

    As Required:

 

      15   B => ~[~A & ~B]

            4 Conclusion, 9

 

      16   [A => ~[~A & ~B]] & [B => ~[~A & ~B]]

            Join, 8, 15

 

      17   ~[~A & ~B]

            Cases, 1, 16

 

'=>'

 

As Required:

 

18   A | B => ~[~A & ~B]

      4 Conclusion, 1

 

   

    '<='

   

    Prove: ~[~A & ~B] => A | B

   

    Suppose...

 

      19   ~[~A & ~B]

            Premise

 

    Two cases to consider:

 

      20   A | ~A

            Or Not

 

         Case 1

        

         Prove: A => A | B

        

         Suppose...

 

            21   A

                  Premise

 

            22   A | B

                  Arb Or, 21

 

    Case 1

   

    As Required:

 

      23   A => A | B

            4 Conclusion, 21

 

         Case 2

        

         Prove: ~A => A | B

        

         Suppose...

 

            24   ~A

                  Premise

 

             Prove: ~~B

            

             Suppose to the contrary...

 

                  25   ~B

                        Premise

 

                  26   ~A & ~B

                        Join, 24, 25

 

             Obtain contradiction...

 

                  27   ~A & ~B & ~[~A & ~B]

                        Join, 26, 19

 

         As Required:

 

            28   ~~B

                  4 Conclusion, 25

 

            29   B

                  Rem DNeg, 28

 

            30   A | B

                  Arb Or, 29

 

    Case 2

   

    As Required:

 

      31   ~A => A | B

            4 Conclusion, 24

 

      32   [A => A | B] & [~A => A | B]

            Join, 23, 31

 

      33   A | B

            Cases, 20, 32

 

'<='

 

As Required:

 

34   ~[~A & ~B] => A | B

      4 Conclusion, 19

 

Joing results...

 

35   [A | B => ~[~A & ~B]] & [~[~A & ~B] => A | B]

      Join, 18, 34

 

As Required:

 

36   A | B <=> ~[~A & ~B]

      Iff-And, 35