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Propositional logic, boolean arithmetic and binary arithmetic

                     Vladimir Cuesta †
                     Instituto de Ciencias Nucleares, Universidad Nacional Aut´ noma de M´ xico, 70-543, Ciudad de M´ xico, M´ xico
                                                                              o          e                          e        e


                     Abstract. I present a set of logic tables or truth tables based only in the negation ¬ and in the conjunction . In fact, I show that the
                     disjunction, the implication, the double implication can be written in terms of the previous one, aditionally defining new operations it is
                     possible to write complete tables as tautologies and anti tautologies. This differs from the NAND and NOR approach.
                          Aditionally, taking a non empty set A and the empty set ∅ it is possible to write different tables using the intersection and complement
                     operations these tables can be though as ”the disjunction”, ”the implication” and ”the double implication”, defining new operations we can
                     obtain complete tables with the set A and the empty set ∅.
                          Finally, I construct based on the binary arithmetic a set of tables based on the operations 1 − r and r · s where r and s belongs to the
                     set {0, 1} in such a way that it enables construct operations that resemble ”the disjunction”, ”the implication” and ”the double implication”
                     in logic tables, with appropiates definitions I obtain complete tables with zeros and ones.




1. Introduction

One branch of mathematics is without doubt discrete mathematics, one of the fundamental notions is that it does not exist
continuity in the structures of all the theories that belongs to it. Inside this branch of mathematics lies the mathematical logic, if
people are very stricted set theory and binary arithmetic associated with computer science can be labelled as discrete mathematics.
     Inside mathematical logic there exists different notions and branches, propositional calculus is one of them, this one is
interested in the determination of truth values of different logical propositions. Inside propositional calculus we can find five
important ingredients the first are the propositions, the second are the different connectives, the third are the quantifiers, the
fourth are the different parenthesis and finally the truth values.
     Different studies are interested in the interpretation of negation, these studies give different interpretation, one of them is
for example to take (¬)n p = p (see [1] for instance), I mean if take a proposition p and we make the negation n times, then the
result is the original one, in this work I will assume that the negation of the negation is the identity.
     In the present written I will study some notions of propositional calculus only, first order logic, second order logic and
higher order logics are far away of my study, including notions as consistence, completeness and so on. In fact there are so many
information in the logic literature (see for example [2] for a discussion about completeness) that is impossible to review all.
     In the present written I will present different tables for logic, sets and binary aritmetic (that resemble the zeros and ones
of computer science), I will compare propositional calculus with the other two branches of mathematics and another important
point is the construction of connectives in a different way like the usual NAND and NOR approach.

2. Usual case

2.1. Logic
Let p and q be two different sentences with truth values, I mean propositions p and q can have the truth values true (T ) or false
(F ). Aditionally, for making effective a propositional calculus we need parenthesis ”(”, ”)” and different connectives like the
conjuction p q, the negation ¬p, the disjunction p q, the implication p ⇒ q, the double implication p ⇔ q, the Sheffer stroke
or the NAND operation p ↑ q, the Pierce arrow or NOR operation p ↓ q and the exclusive disjunction p q.
  †   vladimir.cuesta@nucleares.unam.mx
Propositional logic, boolean arithmetic and binary arithmetic                                                                                 2

     I begin discussing some usual properties of the NAND operation. Let p and q be two propositions with values of truth
T or F , in this case the NAND operation is equivalent to p ↑ q ≡ ¬(p q) with this in mind it is possible to show that
the connectives ¬, , , ⇒, the double implication ⇔, and the exclusive disjunction               and so on can be written in terms
of the NAND operation, the expressions are as follows (altough I show only the expressions for the four first connectives),
¬p ≡ p ↑ p, p q ≡ (p ↑ q) ↑ (p ↑ q), p q ≡ (p ↑ p) ↑ (q ↑ q), p ⇒ q ≡ p ↑ (q ↑ q) ≡ p ↑ (p ↑ q).
     In this paper I follow a different approach, I use the negation ¬ and the conjuction for expressing the connectives ¬, ,
  , ⇒, the double implication ⇔ and so on in terms like I said of the negation and conjuction only, I show the expressions for
the first four connectives, ¬p ≡ ¬p, p q ≡ p q, p q ≡ ¬ (¬p ¬q) , p ⇒ q ≡ (p ¬q) , all the set of tables show
what I said and in fact, It shows a serie of different connectives and the interested reader can compare the different tables for
determining the equivalences (see [4] and [5] for a discussion about logic, sets and different connectives)
                   p    q     ¬p    ¬q     p       q       p⇒q       p       ¬q   ¬ (p       ¬q) p         q    ¬p       ¬q     ¬ (¬p ¬q)
                   T    T     F     F          T            T            F               T             T             F               T
                   T    F     F     T          F            F            T               F             T             F               T
                   F    T     T     F          F            T            F               T             T             F               T
                   F    F     T     T          F            T            F               T             F             T               F

                   q⇒p        q     ¬p    ¬ (q         ¬p) p ⇔ q             (¬ (p   ¬q))        (¬ (p         ¬q)) p         q (¬ (q ⇔ p))
                    T             F              T           T                               T                                    F
                    T             F              T           F                               F                                    T
                    F             T              F           F                               F                                    T
                    T             F              T           T                               T                                    F

                   ¬ ((¬ (p       ¬q)) (¬ (p               ¬q))) p ↑ q         ¬ (p q) p ↓ q             ¬q        ¬q
                                     F                            F                F    F                      F
                                     T                             T               T    F                      F
                                     T                             T               T    F                      F
                                     F                             T               T     T                     T

2.2. Sets
Let A be a non empty set and ∅ the empty set, in boolean arithmetic there exists different operations for sets, the intersection ,
the complement C , the union and connectives like parenthesis ”(”, ”)”. In this point I will do a comparison between logic
operations and boolean arithmetic, in fact we can make the following association the true value T will correspond with the non
empty set A and the false value will correspond with the empty set ∅, the logic operation of negation ¬ will correspond with
the complement operation C in boolean arithmetic and the logic operation of conjuction ¬ will correspond with the intersection
operation in boolean arithmetic (see [4]) in the following sense:
     ♦ The negation of a true value is false, the complement of A is the empty set ∅.
     ♦ The negation of a false value is true, the complement of ∅ is A.
     ♦ The conjuction of two true values is true, the conjuction of two A s is A, the conjuction of true and false values in any
order is false, the conjuction of A and ∅ in any order is the set ∅.
     With this in mind, we show a serie of tables similar to the previous section, with the adecuate comparison,
                                                                                         C                                      C
                   R    S RC         SC     R          S     R   SC          R    SC          RC       SC          RC      SC       S    RC
                   A    A ∅           ∅            A             ∅                A                ∅                     A              ∅
                   A    ∅  ∅         A             ∅             A                ∅                ∅                     A              ∅
                   ∅    A A           ∅            ∅             ∅                A                ∅                     A              A
                   ∅    ∅ A          A             ∅             ∅                A                A                     ∅              ∅
Propositional logic, boolean arithmetic and binary arithmetic                                                                     3

                              C                C                 C             C                C C            C
                     S   RC         R     SC           S    RC        R   SC           S   RC         (R   S)      RC       SC
                         A                         A                               ∅                       ∅            ∅
                         A                         ∅                               A                       A            ∅
                         ∅                         ∅                               A                       A            ∅
                         A                         A                               ∅                       A            A

2.3. Computer science
In the present section of the article I present a serie of tables based on the binary code or an arithmetic of base two (see [6] and
[5] for a discussion of the subject), just for completeness I remember some values of base two 0 = 0 +2 0, 1 = 0 +2 1, 1 =
1 +2 1, 0 = 1 +2 1, in fact the previous table can be generalized in the following way, for example if we take the number 3 this
number can be written as 3 = 2 × 1 + 1 and in this case the residue for this arithmetic of base 2 is 1, and by this reason in this
binary arithmetic 2 × 1 +2 1 = 1 and so on, for example 4 = 2 × 2 + 0 = 2 × 2 +2 0 = 0 just the end.
      Now in a similar way I can make a comparison between proposional logic and binary arithmetic. The logic value T will
correspond with the value 1 in the binary arithmetic (or the aritmetic adopted to the computer science), the logic value F will
correspond with the value 0, the connectives are similar in both cases, I mean the parentheses ”(”, ”)” appears in both branches,
the logic operation of negation ¬ will correspond with 1 − m, where m belongs to the set {0, 1} and the logic conjuction will
correspond with the product operation ·, the previous discussion can be resumed as follows
      ♦ The negation of the true value T is false F , 1 −2 1 = 0,
      ♦ The negation of the false value F is true T 1 −2 0 = 0,
      ♦ The conjuction of two true values T and T is true T , 1 · 1 = 1; the conjuction of one true value T and one false value F
is false F ; the product of 1 and 0 or 0 and 0 in any order is 0.
                   m n       1−m 1−n           m·n         m · (1 − n) 1 − m · (1 − n) (1 − m) · (1 − n)
                   1 1        0   0             1               0            1                0
                   1 0        0   1             0               1            0                0
                   0 1        1   0             0               0            1                0
                   0 0        1   1             0               0            1                1

                   1 − (1 − m) · (1 − n) n · (1 − m) 1 − n · (1 − m) (1 − m · (1 − n)) · (1 − n · (1 − m))
                            1                  0             1                        1
                            1                  0             1                        0
                            1                  1             0                        0
                            0                  0             1                        1

                   1 − (1 − m · (1 − n)) · (1 − n · (1 − m)) 1 − m · n         (1 − m) · (1 − n)
                                      0                          0                    0
                                      1                          1                    0
                                      1                          1                    0
                                      0                          1                    1
Resuming this section I will put the equivalences what I said in the following table
                   P ropositional       logic Boolean        arithmetic Binary arithmetic
                             T                               A                 1
                             F                               ∅                 0
                                                             C
                             ¬                                                1−m
                                                                              m·n
Propositional logic, boolean arithmetic and binary arithmetic                                                                                                                        4

3. Extreme cases I

3.1. Logic
Let p and q two propositions with true value T or false value F , with the help of the negation and the conjuction I define the
following operation p 1 q ≡ ¬ (p (q ¬q)) , this operation defines a complete tautology table. Now proceeding carefully I
show the following set of truth tables
                   p     q         ¬p     ¬q      p       1   q       p   1   ¬q    ¬ (p        1   ¬q) ¬p      1   ¬q       ¬ (¬p 1 ¬q) q 1 ¬p               ¬ (q         1   ¬p)
                   T     T         F      F           T                   T                 F                   T                 F        T                           F
                   T     F         F      T           T                   T                 F                   T                 F        T                           F
                   F     T         T      F           T                   T                 F                   T                 F        T                           F
                   F     F         T      T           T                   T                 F                   T                 F        T                           F

                   (¬ (p       1   ¬q))    1   (¬ (q          1   ¬p)) ¬ ((¬ (p              1      ¬q)) 1 (¬ (q     1   ¬p))) ¬ (p 1 q) ¬p 1 ¬q
                                          T                                                             F                          F       T
                                          T                                                             F                          F       T
                                          T                                                             F                          F       T
                                          T                                                             F                          F       T

3.2. Sets
Let R and S two sets with two possible values one of them would be a non empty set A and the empty set ∅. With the help of
                                                                                                C
the complement operation and the intersection I define the following R 1 S ≡ R          S SC        , this operation defines a table
with the element the empty set ∅, making a straighforward calculation I obtain the following tables
                                                                                                            C                                    C
                   R     S RC             SC       R          1   S       R   1    SC       R       1 SC        RC       1   SC    RC   1   SC        S       1   RC
                   A     A ∅               ∅              A                   A                      ∅               A                  ∅                 A
                   A     ∅  ∅             A               A                   A                      ∅               A                  ∅                 A
                   ∅     A A               ∅              A                   A                      ∅               A                  ∅                 A
                   ∅     ∅ A              A               A                   A                      ∅               A                  ∅                 A

                                    C                             C                                 C                        C C             C
                     S     1   RC             R       1   SC              1        R    1   SC          1   S   1   RC             (R   1   S)       RC   1   SC
                                                                      C
                                               S          1RC
                           ∅                              A                                             A                               ∅                 A
                           ∅                              A                                             A                               ∅                 A
                           ∅                              A                                             A                               ∅                 A
                           ∅                              A                                             A                               ∅                 A

3.3. Computer science
Let m, n two numbers with values 0 and 1 and the operation 1 − m for example and the product of m and n (m · n), with results
evaluated in the binary arithmetic I make the definition m ∗1 n ≡ 1 − n · (m · (1 − m))) , this produces results only with the
number one, for complementing the present subsection I show a serie of tables for a lot of computacional operations,
                   m n          1−m 1−n                        m ∗1 n          m ∗1 (1 − n) 1 − m ∗1 (1 − n) (1 − m) ∗1 (1 − n)
                   1 1           0   0                           1                  1              0                 1
                   1 0           0   1                           1                  1              0                 1
                   0 1           1   0                           1                  1              0                 1
                   0 0           1   1                           1                  1              0                 1
Propositional logic, boolean arithmetic and binary arithmetic                                                                                                                                           5

                  1 − (1 − m) ∗1 (1 − n) n ∗1 (1 − m) 1 − n ∗1 (1 − m) (1 − m ∗1 (1 − n)) ∗1 (1 − n ∗1 (1 − m))
                            0                  1              0                           1
                            0                  1              0                           1
                            0                  1              0                           1
                            0                  1              0                           1

                  1 − (1 − m ∗1 (1 − n)) ∗1 (1 − n ∗1 (1 − m)) 1 − m ∗1 n                                                     (1 − m) ∗1 (1 − n)
                                       0                           0                                                                  1
                                       0                           0                                                                  1
                                       0                           0                                                                  1
                                       0                           0                                                                  1

4. Extreme cases II

4.1. Logic
Let p and q two propositions with true T or false F truth values, I impose the following definition p                                                               2q   ≡p           (q   ¬q) , the
previous definition produces a table full of false values F .
     With the help of this logic operation I find the following set of tables
                  p       q       ¬p     ¬q      p       2   q        p      2   ¬q    ¬ (p        2   ¬q) ¬p            2   ¬q       ¬ (¬p       2   ¬q) q         2   ¬p       ¬ (q         2   ¬p)
                  T       T       F      F           F                       F                 T                       F                      T                     F                     T
                  T       F       F      T           F                       F                 T                       F                      T                     F                     T
                  F       T       T      F           F                       F                 T                       F                      T                     F                     T
                  F       F       T      T           F                       F                 T                       F                      T                     F                     T

                  (¬ (p       2   ¬q))    2   (¬ (q              2   ¬p)) ¬ ((¬ (p              2      ¬q))       2   (¬ (q   2   ¬p))) ¬ (p               2   q) ¬p         2   ¬q
                                         F                                                                    T                                        T                    F
                                         F                                                                    T                                        T                    F
                                         F                                                                    T                                        T                    F
                                         F                                                                    T                                        T                    F

4.2. Sets
Let A a non empty set and ∅ the empty set, I define the operation R                                         2S     ≡R          S        S C , this operation produces a table that
it is full of the empty set ∅.
                                                                                                                  C                                                C
                  R       S RC           SC       R          2   S           R   2    SC       R       2 SC             RC        2   SC      RC       2   SC           S        2   RC
                  A       A ∅             ∅                  ∅                   ∅                      A                     ∅                        A                         ∅
                  A       ∅  ∅           A                   ∅                   ∅                      A                     ∅                        A                         ∅
                  ∅       A A             ∅                  ∅                   ∅                      A                     ∅                        A                         ∅
                  ∅       ∅ A            A                   ∅                   ∅                      A                     ∅                        A                         ∅

                                   C                                 C                                 C                              C C                      C
                      S   2   RC             R       2   SC                  2        R    2   SC             2   S      2   RC               (R       2   S)          RC       2   SC
                                                                         C
                                              S          2RC
                          A                              ∅                                                    ∅                                       A                      ∅
                          A                              ∅                                                    ∅                                       A                      ∅
                          A                              ∅                                                    ∅                                       A                      ∅
                          A                              ∅                                                    ∅                                       A                      ∅
Propositional logic, boolean arithmetic and binary arithmetic                                                                                      6

4.3. Computer science
Let m, n two numbers taking values 0 and 1 with the operation 1 − m and the product of m and n, I mean m · n, the result will
be into the same set of values 0 and 1, I make the following definition m ∗2 n ≡ n · (m · (1 − m)) this definition produces results
only with the number zero.
     For finishing I show a serie of tables for different computacional operations,
                      m n       1−m 1−n             m ∗1 n     m ∗1 (1 − n) 1 − m ∗1 (1 − n) (1 − m) ∗1 (1 − n)
                      1 1        0   0                0             0              1                 0
                      1 0        0   1                0             0              1                 0
                      0 1        1   0                0             0              1                 0
                      0 0        1   1                0             0              1                 0

                      1 − (1 − m) ∗1 (1 − n) n ∗1 (1 − m) 1 − n ∗1 (1 − m) (1 − m ∗1 (1 − n)) ∗1 (1 − n ∗1 (1 − m))
                                1                  0              1                           0
                                1                  0              1                           0
                                1                  0              1                           0
                                1                  0              1                           0

                      1 − (1 − m ∗1 (1 − n)) ∗1 (1 − n ∗1 (1 − m)) 1 − m ∗1 n                   (1 − m) ∗1 (1 − n)
                                           1                           1                                0
                                           1                           1                                0
                                           1                           1                                0
                                           1                           1                                0

5. Conclusions and perspectives

In the article I have presented a different construction for the connectives used with the help of the NAND and NOR operations
for propositional logic, boolean arithmetic and binary arithmetic (or arithmetic for the computer science), all what I make is
to base my work in representing different connectives with the negation and conjuction; the complement and intersection and
1 − ( ) and product respectively. As a second point altough probably there are more that those, I have defined as first example
a 3-nuple of operations ( 1 ,        1,   ∗1 ) and as a second example a 3-nuple of operations ( 2 ,         2,    ∗2 ). However, for
finishing my work I present another possible 3-nuple ( 3 ,          3 , ∗3 ), the question that it remains open is the following: Do it
exist a explicit representation for this 3-nuple in terms of the the negation and conjuction; the complement and intersection and
1 − ( ) and product?
                      p    q    p    3   q         R    S R 3S                   m n       m ∗3 n
                      T    T        T              A    A  A                     1 1         1
                      T    F        F              A    ∅   ∅                    1 0         0
                      F    T        T              ∅    A  A                     0 1         1
                      F    F        F              ∅    ∅   ∅                    0 0         0
Probably the answer is in the affirmative, but the question remains open at least by myself.

References
[1] Tzouvaras, Athanassios, Periodicity of negation, Notra Dame Journal of Formal Logic, 42, 2 (2001),
[2] McCarty, D. C., Intuitionistic Completeness and Classical Logic, Notra Dame Journal of Formal Logic, 43, 4 (2002),
[3] Garbacz, Pawel, Logics of relative Identity Notra Dame Journal of Formal Logic, 43, 1 (2002),
[4] Gindikin, S. G., Algebraic Logic, University of Moscow, Springer-Verlag (1985),
[5] Harris, John W., Stocker Horst, Handbook of mathematics and Computacional Science, New Haven, Connecticut and Frankfurt, Germany, Springer-Verlag
       (1998),
[6] Manna, Zohar, Mathematical Theory of computation Stanford University, Dover (1974).

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Propositional Logic, Boolean Arithmetic And Binary Arithmetic

  • 1. Propositional logic, boolean arithmetic and binary arithmetic Vladimir Cuesta † Instituto de Ciencias Nucleares, Universidad Nacional Aut´ noma de M´ xico, 70-543, Ciudad de M´ xico, M´ xico o e e e Abstract. I present a set of logic tables or truth tables based only in the negation ¬ and in the conjunction . In fact, I show that the disjunction, the implication, the double implication can be written in terms of the previous one, aditionally defining new operations it is possible to write complete tables as tautologies and anti tautologies. This differs from the NAND and NOR approach. Aditionally, taking a non empty set A and the empty set ∅ it is possible to write different tables using the intersection and complement operations these tables can be though as ”the disjunction”, ”the implication” and ”the double implication”, defining new operations we can obtain complete tables with the set A and the empty set ∅. Finally, I construct based on the binary arithmetic a set of tables based on the operations 1 − r and r · s where r and s belongs to the set {0, 1} in such a way that it enables construct operations that resemble ”the disjunction”, ”the implication” and ”the double implication” in logic tables, with appropiates definitions I obtain complete tables with zeros and ones. 1. Introduction One branch of mathematics is without doubt discrete mathematics, one of the fundamental notions is that it does not exist continuity in the structures of all the theories that belongs to it. Inside this branch of mathematics lies the mathematical logic, if people are very stricted set theory and binary arithmetic associated with computer science can be labelled as discrete mathematics. Inside mathematical logic there exists different notions and branches, propositional calculus is one of them, this one is interested in the determination of truth values of different logical propositions. Inside propositional calculus we can find five important ingredients the first are the propositions, the second are the different connectives, the third are the quantifiers, the fourth are the different parenthesis and finally the truth values. Different studies are interested in the interpretation of negation, these studies give different interpretation, one of them is for example to take (¬)n p = p (see [1] for instance), I mean if take a proposition p and we make the negation n times, then the result is the original one, in this work I will assume that the negation of the negation is the identity. In the present written I will study some notions of propositional calculus only, first order logic, second order logic and higher order logics are far away of my study, including notions as consistence, completeness and so on. In fact there are so many information in the logic literature (see for example [2] for a discussion about completeness) that is impossible to review all. In the present written I will present different tables for logic, sets and binary aritmetic (that resemble the zeros and ones of computer science), I will compare propositional calculus with the other two branches of mathematics and another important point is the construction of connectives in a different way like the usual NAND and NOR approach. 2. Usual case 2.1. Logic Let p and q be two different sentences with truth values, I mean propositions p and q can have the truth values true (T ) or false (F ). Aditionally, for making effective a propositional calculus we need parenthesis ”(”, ”)” and different connectives like the conjuction p q, the negation ¬p, the disjunction p q, the implication p ⇒ q, the double implication p ⇔ q, the Sheffer stroke or the NAND operation p ↑ q, the Pierce arrow or NOR operation p ↓ q and the exclusive disjunction p q. † vladimir.cuesta@nucleares.unam.mx
  • 2. Propositional logic, boolean arithmetic and binary arithmetic 2 I begin discussing some usual properties of the NAND operation. Let p and q be two propositions with values of truth T or F , in this case the NAND operation is equivalent to p ↑ q ≡ ¬(p q) with this in mind it is possible to show that the connectives ¬, , , ⇒, the double implication ⇔, and the exclusive disjunction and so on can be written in terms of the NAND operation, the expressions are as follows (altough I show only the expressions for the four first connectives), ¬p ≡ p ↑ p, p q ≡ (p ↑ q) ↑ (p ↑ q), p q ≡ (p ↑ p) ↑ (q ↑ q), p ⇒ q ≡ p ↑ (q ↑ q) ≡ p ↑ (p ↑ q). In this paper I follow a different approach, I use the negation ¬ and the conjuction for expressing the connectives ¬, , , ⇒, the double implication ⇔ and so on in terms like I said of the negation and conjuction only, I show the expressions for the first four connectives, ¬p ≡ ¬p, p q ≡ p q, p q ≡ ¬ (¬p ¬q) , p ⇒ q ≡ (p ¬q) , all the set of tables show what I said and in fact, It shows a serie of different connectives and the interested reader can compare the different tables for determining the equivalences (see [4] and [5] for a discussion about logic, sets and different connectives) p q ¬p ¬q p q p⇒q p ¬q ¬ (p ¬q) p q ¬p ¬q ¬ (¬p ¬q) T T F F T T F T T F T T F F T F F T F T F T F T T F F T F T T F T F F T T F T F T F T F q⇒p q ¬p ¬ (q ¬p) p ⇔ q (¬ (p ¬q)) (¬ (p ¬q)) p q (¬ (q ⇔ p)) T F T T T F T F T F F T F T F F F T T F T T T F ¬ ((¬ (p ¬q)) (¬ (p ¬q))) p ↑ q ¬ (p q) p ↓ q ¬q ¬q F F F F F T T T F F T T T F F F T T T T 2.2. Sets Let A be a non empty set and ∅ the empty set, in boolean arithmetic there exists different operations for sets, the intersection , the complement C , the union and connectives like parenthesis ”(”, ”)”. In this point I will do a comparison between logic operations and boolean arithmetic, in fact we can make the following association the true value T will correspond with the non empty set A and the false value will correspond with the empty set ∅, the logic operation of negation ¬ will correspond with the complement operation C in boolean arithmetic and the logic operation of conjuction ¬ will correspond with the intersection operation in boolean arithmetic (see [4]) in the following sense: ♦ The negation of a true value is false, the complement of A is the empty set ∅. ♦ The negation of a false value is true, the complement of ∅ is A. ♦ The conjuction of two true values is true, the conjuction of two A s is A, the conjuction of true and false values in any order is false, the conjuction of A and ∅ in any order is the set ∅. With this in mind, we show a serie of tables similar to the previous section, with the adecuate comparison, C C R S RC SC R S R SC R SC RC SC RC SC S RC A A ∅ ∅ A ∅ A ∅ A ∅ A ∅ ∅ A ∅ A ∅ ∅ A ∅ ∅ A A ∅ ∅ ∅ A ∅ A A ∅ ∅ A A ∅ ∅ A A ∅ ∅
  • 3. Propositional logic, boolean arithmetic and binary arithmetic 3 C C C C C C C S RC R SC S RC R SC S RC (R S) RC SC A A ∅ ∅ ∅ A ∅ A A ∅ ∅ ∅ A A ∅ A A ∅ A A 2.3. Computer science In the present section of the article I present a serie of tables based on the binary code or an arithmetic of base two (see [6] and [5] for a discussion of the subject), just for completeness I remember some values of base two 0 = 0 +2 0, 1 = 0 +2 1, 1 = 1 +2 1, 0 = 1 +2 1, in fact the previous table can be generalized in the following way, for example if we take the number 3 this number can be written as 3 = 2 × 1 + 1 and in this case the residue for this arithmetic of base 2 is 1, and by this reason in this binary arithmetic 2 × 1 +2 1 = 1 and so on, for example 4 = 2 × 2 + 0 = 2 × 2 +2 0 = 0 just the end. Now in a similar way I can make a comparison between proposional logic and binary arithmetic. The logic value T will correspond with the value 1 in the binary arithmetic (or the aritmetic adopted to the computer science), the logic value F will correspond with the value 0, the connectives are similar in both cases, I mean the parentheses ”(”, ”)” appears in both branches, the logic operation of negation ¬ will correspond with 1 − m, where m belongs to the set {0, 1} and the logic conjuction will correspond with the product operation ·, the previous discussion can be resumed as follows ♦ The negation of the true value T is false F , 1 −2 1 = 0, ♦ The negation of the false value F is true T 1 −2 0 = 0, ♦ The conjuction of two true values T and T is true T , 1 · 1 = 1; the conjuction of one true value T and one false value F is false F ; the product of 1 and 0 or 0 and 0 in any order is 0. m n 1−m 1−n m·n m · (1 − n) 1 − m · (1 − n) (1 − m) · (1 − n) 1 1 0 0 1 0 1 0 1 0 0 1 0 1 0 0 0 1 1 0 0 0 1 0 0 0 1 1 0 0 1 1 1 − (1 − m) · (1 − n) n · (1 − m) 1 − n · (1 − m) (1 − m · (1 − n)) · (1 − n · (1 − m)) 1 0 1 1 1 0 1 0 1 1 0 0 0 0 1 1 1 − (1 − m · (1 − n)) · (1 − n · (1 − m)) 1 − m · n (1 − m) · (1 − n) 0 0 0 1 1 0 1 1 0 0 1 1 Resuming this section I will put the equivalences what I said in the following table P ropositional logic Boolean arithmetic Binary arithmetic T A 1 F ∅ 0 C ¬ 1−m m·n
  • 4. Propositional logic, boolean arithmetic and binary arithmetic 4 3. Extreme cases I 3.1. Logic Let p and q two propositions with true value T or false value F , with the help of the negation and the conjuction I define the following operation p 1 q ≡ ¬ (p (q ¬q)) , this operation defines a complete tautology table. Now proceeding carefully I show the following set of truth tables p q ¬p ¬q p 1 q p 1 ¬q ¬ (p 1 ¬q) ¬p 1 ¬q ¬ (¬p 1 ¬q) q 1 ¬p ¬ (q 1 ¬p) T T F F T T F T F T F T F F T T T F T F T F F T T F T T F T F T F F F T T T T F T F T F (¬ (p 1 ¬q)) 1 (¬ (q 1 ¬p)) ¬ ((¬ (p 1 ¬q)) 1 (¬ (q 1 ¬p))) ¬ (p 1 q) ¬p 1 ¬q T F F T T F F T T F F T T F F T 3.2. Sets Let R and S two sets with two possible values one of them would be a non empty set A and the empty set ∅. With the help of C the complement operation and the intersection I define the following R 1 S ≡ R S SC , this operation defines a table with the element the empty set ∅, making a straighforward calculation I obtain the following tables C C R S RC SC R 1 S R 1 SC R 1 SC RC 1 SC RC 1 SC S 1 RC A A ∅ ∅ A A ∅ A ∅ A A ∅ ∅ A A A ∅ A ∅ A ∅ A A ∅ A A ∅ A ∅ A ∅ ∅ A A A A ∅ A ∅ A C C C C C C S 1 RC R 1 SC 1 R 1 SC 1 S 1 RC (R 1 S) RC 1 SC C S 1RC ∅ A A ∅ A ∅ A A ∅ A ∅ A A ∅ A ∅ A A ∅ A 3.3. Computer science Let m, n two numbers with values 0 and 1 and the operation 1 − m for example and the product of m and n (m · n), with results evaluated in the binary arithmetic I make the definition m ∗1 n ≡ 1 − n · (m · (1 − m))) , this produces results only with the number one, for complementing the present subsection I show a serie of tables for a lot of computacional operations, m n 1−m 1−n m ∗1 n m ∗1 (1 − n) 1 − m ∗1 (1 − n) (1 − m) ∗1 (1 − n) 1 1 0 0 1 1 0 1 1 0 0 1 1 1 0 1 0 1 1 0 1 1 0 1 0 0 1 1 1 1 0 1
  • 5. Propositional logic, boolean arithmetic and binary arithmetic 5 1 − (1 − m) ∗1 (1 − n) n ∗1 (1 − m) 1 − n ∗1 (1 − m) (1 − m ∗1 (1 − n)) ∗1 (1 − n ∗1 (1 − m)) 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 − (1 − m ∗1 (1 − n)) ∗1 (1 − n ∗1 (1 − m)) 1 − m ∗1 n (1 − m) ∗1 (1 − n) 0 0 1 0 0 1 0 0 1 0 0 1 4. Extreme cases II 4.1. Logic Let p and q two propositions with true T or false F truth values, I impose the following definition p 2q ≡p (q ¬q) , the previous definition produces a table full of false values F . With the help of this logic operation I find the following set of tables p q ¬p ¬q p 2 q p 2 ¬q ¬ (p 2 ¬q) ¬p 2 ¬q ¬ (¬p 2 ¬q) q 2 ¬p ¬ (q 2 ¬p) T T F F F F T F T F T T F F T F F T F T F T F T T F F F T F T F T F F T T F F T F T F T (¬ (p 2 ¬q)) 2 (¬ (q 2 ¬p)) ¬ ((¬ (p 2 ¬q)) 2 (¬ (q 2 ¬p))) ¬ (p 2 q) ¬p 2 ¬q F T T F F T T F F T T F F T T F 4.2. Sets Let A a non empty set and ∅ the empty set, I define the operation R 2S ≡R S S C , this operation produces a table that it is full of the empty set ∅. C C R S RC SC R 2 S R 2 SC R 2 SC RC 2 SC RC 2 SC S 2 RC A A ∅ ∅ ∅ ∅ A ∅ A ∅ A ∅ ∅ A ∅ ∅ A ∅ A ∅ ∅ A A ∅ ∅ ∅ A ∅ A ∅ ∅ ∅ A A ∅ ∅ A ∅ A ∅ C C C C C C S 2 RC R 2 SC 2 R 2 SC 2 S 2 RC (R 2 S) RC 2 SC C S 2RC A ∅ ∅ A ∅ A ∅ ∅ A ∅ A ∅ ∅ A ∅ A ∅ ∅ A ∅
  • 6. Propositional logic, boolean arithmetic and binary arithmetic 6 4.3. Computer science Let m, n two numbers taking values 0 and 1 with the operation 1 − m and the product of m and n, I mean m · n, the result will be into the same set of values 0 and 1, I make the following definition m ∗2 n ≡ n · (m · (1 − m)) this definition produces results only with the number zero. For finishing I show a serie of tables for different computacional operations, m n 1−m 1−n m ∗1 n m ∗1 (1 − n) 1 − m ∗1 (1 − n) (1 − m) ∗1 (1 − n) 1 1 0 0 0 0 1 0 1 0 0 1 0 0 1 0 0 1 1 0 0 0 1 0 0 0 1 1 0 0 1 0 1 − (1 − m) ∗1 (1 − n) n ∗1 (1 − m) 1 − n ∗1 (1 − m) (1 − m ∗1 (1 − n)) ∗1 (1 − n ∗1 (1 − m)) 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 − (1 − m ∗1 (1 − n)) ∗1 (1 − n ∗1 (1 − m)) 1 − m ∗1 n (1 − m) ∗1 (1 − n) 1 1 0 1 1 0 1 1 0 1 1 0 5. Conclusions and perspectives In the article I have presented a different construction for the connectives used with the help of the NAND and NOR operations for propositional logic, boolean arithmetic and binary arithmetic (or arithmetic for the computer science), all what I make is to base my work in representing different connectives with the negation and conjuction; the complement and intersection and 1 − ( ) and product respectively. As a second point altough probably there are more that those, I have defined as first example a 3-nuple of operations ( 1 , 1, ∗1 ) and as a second example a 3-nuple of operations ( 2 , 2, ∗2 ). However, for finishing my work I present another possible 3-nuple ( 3 , 3 , ∗3 ), the question that it remains open is the following: Do it exist a explicit representation for this 3-nuple in terms of the the negation and conjuction; the complement and intersection and 1 − ( ) and product? p q p 3 q R S R 3S m n m ∗3 n T T T A A A 1 1 1 T F F A ∅ ∅ 1 0 0 F T T ∅ A A 0 1 1 F F F ∅ ∅ ∅ 0 0 0 Probably the answer is in the affirmative, but the question remains open at least by myself. References [1] Tzouvaras, Athanassios, Periodicity of negation, Notra Dame Journal of Formal Logic, 42, 2 (2001), [2] McCarty, D. C., Intuitionistic Completeness and Classical Logic, Notra Dame Journal of Formal Logic, 43, 4 (2002), [3] Garbacz, Pawel, Logics of relative Identity Notra Dame Journal of Formal Logic, 43, 1 (2002), [4] Gindikin, S. G., Algebraic Logic, University of Moscow, Springer-Verlag (1985), [5] Harris, John W., Stocker Horst, Handbook of mathematics and Computacional Science, New Haven, Connecticut and Frankfurt, Germany, Springer-Verlag (1998), [6] Manna, Zohar, Mathematical Theory of computation Stanford University, Dover (1974).