Conjunction/disjunction duality explained

In propositional logic and Boolean algebra, there is a duality between conjunction and disjunction,[1] [2] [3] also called the duality principle.[4] [5] [6] It is, undoubtedly, the most widely known example of duality in logic. The duality consists in these metalogical theorems:

\varphiD

is used as notation to designate the result of replacing every instance of conjunction with disjunction, and every instance of disjunction with conjunction (e.g.

p\landq

with

q\lorp

, or vice-versa), in a given formula

\varphi

, and if

\overline{\varphi}

is used as notation for replacing every sentence-letter in

\varphi

with its negation (e.g.,

p

with

\negp

), and if the symbol

\models

is used for semantic consequence and ⟚ for semantical equivalence between logical formulas, then it is demonstrable that

\varphiD

 ⟚ 

\neg\overline{\varphi}

,[7] and also that

\varphi\models\psi

if, and only if,

\psiD\models\varphiD

, and furthermore that if

\varphi

 ⟚ 

\psi

then

\varphiD

 ⟚ 

\psiD

. (In this context,

\overline{\varphi}D

is called the dual of a formula

\varphi

.)

This article will prove these results, in the and sections respectively.

Mutual definability

Because of their semantics, i.e. the way they are commonly interpreted in classical propositional logic, conjunction and disjunction can be defined in terms of each other with the aid of negation, so that consequently, only one of them needs to be taken as primitive. For example, if conjunction (∧) and negation (¬) are taken as primitives, then disjunction (∨) can be defined as follows:[8]

\varphi\lor\psi:\equiv\neg(\neg\varphi\land\neg\psi).

(1)

Alternatively, if disjunction is taken as primitive, then conjunction can be defined as follows:[9]

\varphi\land\psi:\equiv\neg(\neg\varphi\lor\neg\psi).

(2)

Also, each of these equivalences can be derived from the other one; for example, if (1) is taken as primitive, then (2) is obtained as follows:

\neg(\neg\varphi\lor\neg\psi)\equiv\neg\neg(\neg\varphi\land\neg\psi)\equiv\varphi\land\psi.

(3)

Functional completeness

Since the Disjunctive Normal Form Theorem shows that the set of connectives

\{\land,\lor,\neg\}

is functionally complete, these results show that the sets of connectives

\{\land,\neg\}

and

\{\lor,\neg\}

are themselves functionally complete as well.

De Morgan's laws

De Morgan's laws also follow from the definitions of these connectives in terms of each other, whichever direction is taken to do it. If conjunction is taken as primitive, then (4) follows immediately from (1), while (5) follows from (1) via (3):

\neg(\varphi\lor\psi)\equiv\neg\varphi\land\neg\psi.

(4)

\neg(\varphi\land\psi)\equiv\neg\varphi\lor\neg\psi.

(5)

Negation is semantically equivalent to dual

Theorem: Let

X

be any sentence in

l{L}[A1,\ldots,An;\land,\lor,\neg]

. (That is, the language with the propositional variables

A1,\ldots,An

and the connectives

\{\land,\lor,\neg\}

.) Let

\overline{X}D

be obtained from

X

by replacing every occurrence of

\land

in

X

by

\lor

, every occurrence of

\lor

by

\land

, and every occurrence of

Ai

by

\negAi

. Then

X

\neg\overline{X}D

. (

\overline{X}D

is called the dual of

X

.)

Proof: A sentence

X

of

l{L}

, where

l{L}

is as in the theorem, will be said to have the property

P

if

X

\neg\overline{X}D

. We shall prove by induction on immediate predecessors that all sentences of

l{L}

have

P

. (An immediate predecessor of a well-formed formula is any of the formulas that are connected by its ; it follows that sentence letters have no immediate predecessors.) So we have to establish that the following two conditions are satisfied: (1) each

Ai

has

P

; and (2) for any non-atomic

X

, from the inductive hypothesis that the immediate predecessors of

X

have

P

, it follows that

X

does also.

  1. Each

    Ai

    clearly has no occurrence of

    \lor

    or

    \land

    , and so
    D
    \overline{A
    i}
    is just

    \negAi

    . So showing that

    Ai

    has

    P

    merely requires showing that

    Ai\Leftrightarrow\neg\negAi

    , which we know to be the case.
  2. The induction step is an argument by cases. If

    X

    is not an

    Ai

    then

    X

    must have one of the following three forms: (i)

    X=Y\lorZ

    , (ii)

    X=Y\landZ

    , or (iii)

    X=\negY

    where

    Y

    and

    Z

    are sentences of

    l{L}

    . If

    X

    is of the form (i) or (ii) it has as immediate predecessors

    Y

    and

    Z

    , while if it is of the form (iii) it has the one immediate predecessor

    Y

    . We shall check that the induction step holds in each of the cases.

    1. Suppose that

      Y

      and

      Z

      each have

      P

      , i.e. that

      Y

      \neg\overline{Y}D

      and

      Z

      \neg\overline{Z}D

      . This supposition, recall, is the inductive hypothesis. From this we infer that

      Y\lorZ

      \neg\overline{Y}D\lor\neg\overline{Z}D

      . By de Morgan's Laws

      \neg\overline{Y}D\land\neg\overline{Z}D

      (\overline{Y}D\land\overline{Z}D)

      . But

      \overline{Y}D\land\overline{Z}D=\overline{(Y\lorZ)}D

      , and

      Y\lorZ=X

      . So it has been shown that the inductive hypothesis implies that

      X

      \neg\overline{X}D

      , i.e.

      X

      has

      P

      as required.
    2. We have the same inductive hypothesis as in (i). So again

      Y

      \neg\overline{Y}D

      and

      Z

      \neg\overline{Z}D

      . Hence

      Y\landZ\Leftrightarrow\neg\overline{Y}D\land\neg\overline{Z}D

      . By de Morgan again,

      \neg\overline{Y}D\land\neg\overline{Z}D

      \neg(\overline{Y}D\land\overline{Z}D)

      . But

      \overline{Y}D\lor\overline{Z}D=\overline{(Y\landZ)}D=\overline{X}D

      . So

      X

      \neg\overline{X}D

      in this case too.
    3. Here the inductive hypothesis is simply that

      Y

      \neg\overline{Y}D

      . Hence

      \negY

      \neg\neg\overline{Y}D

      . But

      \neg\overline{Y}D=\overline{\negY}D=\overline{X}D

      . Hence

      X

      \overline{X}D

      . Q.E.D.

Further duality theorems

Assume

\phi\models\psi

. Then

\overline{\phi}\models\overline{\psi}

by uniform substitution of

\negPi

for

Pi

. Hence,

\neg\psi\models\neg\phi

, by contraposition; so finally,

\psiD\models\phiD

, by the property that

\varphiD

 ⟚ 

\neg\overline{\varphi}

, which was just proved above. And since

\varphiDD=\phi

, it is also true that

\varphi\models\psi

if, and only if,

\psiD\models\phiD

. And it follows, as a corollary, that if

\phi\models\neg\psi

, then

\phiD\models\neg\psiD

.

Conjunctive and disjunctive normal forms

See main article: article, Conjunctive normal form and Disjunctive normal form. For a formula

\varphi

in disjunctive normal form, the formula

\overline{\varphi}D

will be in conjunctive normal form, and given the result that, it will be semantically equivalent to

\neg\varphi

.[10] [11] This provides a procedure for converting between conjunctive normal form and disjunctive normal form.[12] Since the Disjunctive Normal Form Theorem shows that every formula of propositional logic is expressible in disjunctive normal form, every formula is also expressible in conjunctive normal form by means of effecting the conversion to its dual.

Notes and References

  1. Web site: Duality in Logic and Language Internet Encyclopedia of Philosophy . 2024-06-10 . en-US.
  2. Web site: 1.1 Logical Operations . 2024-06-10 . www.whitman.edu.
  3. Book: Look, Brandon C. . The Bloomsbury Companion to Leibniz . 2014-09-25 . Bloomsbury Publishing . 978-1-4725-2485-0 . 127 . en.
  4. Book: Howson, Colin . Logic with trees: an introduction to symbolic logic . 1997 . Routledge . 978-0-415-13342-5 . London ; New York . 41,44–45.
  5. Web site: Boolean algebra, Part 1 Review ICS 241 . 2024-06-10 . courses.ics.hawaii.edu.
  6. Book: Kurki-Suonio, R. . A Practical Theory of Reactive Systems: Incremental Modeling of Dynamic Behaviors . 2005-07-20 . Springer Science & Business Media . 978-3-540-27348-6 . 80–81 . en.
  7. Book: Bostock, David . Intermediate logic . 1997 . Clarendon Press ; Oxford University Press . 978-0-19-875141-0 . Oxford : New York . 62–65.
  8. Book: Makridis, Odysseus . Symbolic logic . 2022 . Palgrave Macmillan . 978-3-030-67395-6 . Palgrave philosophy today . Cham, Switzerland . 133.
  9. Book: Lyons, John . Semantics: Volume 1 . 1977-06-02 . Cambridge University Press . 978-0-521-29165-1 . 145 . en.
  10. Book: Robinson . Alan J. A. . Handbook of Automated Reasoning . Voronkov . Andrei . 2001-06-21 . Gulf Professional Publishing . 978-0-444-82949-8 . 306 . en.
  11. Book: Polkowski, Lech T. . Logic: Reference Book for Computer Scientists: The 2nd Revised, Modified, and Enlarged Edition of "Logics for Computer and Data Sciences, and Artificial Intelligence" . 2023-10-03 . Springer Nature . 978-3-031-42034-4 . 70 . en.
  12. Book: Bagdasar, Ovidiu . Concise Computer Mathematics: Tutorials on Theory and Problems . 2013-10-28 . Springer Science & Business Media . 978-3-319-01751-8 . 36 . en.