Conjunction/disjunction duality

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


 * In classical propositional logic, the connectives for conjunction and disjunction can be defined in terms of each other, and consequently, only one of them needs to be taken as primitive.
 * If $$\varphi^{D}$$ 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. $$p \land q$$ with $$q \lor p$$, 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 $$\neg p$$), and if the symbol $$\models$$ is used for semantic consequence and ⟚ for semantical equivalence between logical formulas, then it is demonstrable that $$\varphi^{D}$$ ⟚ $$\neg \overline{\varphi}$$, and also that $$\varphi \models \psi$$ if, and only if, $$\psi^{D} \models \varphi^{D}$$, and furthermore that if $$\varphi$$ ⟚ $$\psi$$ then $$\varphi^{D}$$ ⟚ $$\psi^{D}$$. (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:


 * $$\varphi \lor \psi :\equiv \neg (\neg \varphi \land \neg \psi).$$ (1)

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


 * $$\varphi \land \psi :\equiv \neg (\neg \varphi \lor \neg \psi).$$ (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 $$\mathcal{L}[A_1, \ldots, A_n; \land, \lor, \neg]$$. (That is, the language with the propositional variables $$A_1, \ldots, A_n$$ 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 $$A_i$$ by $$\neg A_i$$. Then $$X$$ ⟚ $$\neg \overline{X}^{D}$$. ($$\overline{X}^{D}$$ is called the dual of $$X$$.)

Proof:  A sentence $$X$$ of $$\mathcal{L}$$, where $$\mathcal{L}$$ 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 $$\mathcal{L}$$ 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 $$A_i$$ 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.

 Each $$A_i$$ clearly has no occurrence of $$\lor$$ or $$\land$$, and so $$\overline{A_i}^{D}$$ is just $$\neg A_i$$. So showing that $$A_i$$ has $$P$$ merely requires showing that $$A_i \Leftrightarrow \neg \neg A_i$$, which we know to be the case.

The induction step is an argument by cases. If $$X$$ is not an $$A_i$$ then $$X$$ must have one of the following three forms: (i) $$X = Y \lor Z$$, (ii) $$X = Y \land Z$$, or (iii) $$X = \neg Y$$ where $$Y$$ and $$Z$$ are sentences of $$\mathcal{L}$$. 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.

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 $$Y \lor Z$$ ⟚ $$ \neg \overline{Y}^{D} \lor \neg \overline{Z}^{D}$$. By de Morgan's Laws $$\neg \overline{Y}^{D} \land \neg \overline{Z}^{D} $$ ⟚ $$(\overline{Y}^{D} \land \overline{Z}^{D})$$. But $$\overline{Y}^{D} \land \overline{Z}^{D} = \overline{(Y \lor Z)}^{D}$$, and $$Y \lor Z = X$$. So it has been shown that the inductive hypothesis implies that $$X$$ ⟚ $$\neg \overline{X}^{D}$$, i.e. $$X$$ has $$P$$ as required.

We have the same inductive hypothesis as in (i). So again $$Y$$ ⟚ $$\neg \overline{Y}^{D}$$ and $$Z$$ ⟚ $$\neg \overline{Z}^{D}$$. Hence $$Y \land Z \Leftrightarrow \neg \overline{Y}^{D} \land \neg \overline{Z}^{D}$$. By de Morgan again, $$\neg \overline{Y}^{D} \land \neg \overline{Z}^{D}$$ ⟚ $$\neg (\overline{Y}^{D} \land \overline{Z}^{D})$$. But $$\overline{Y}^{D} \lor \overline{Z}^{D} = \overline{(Y \land Z)}^{D} = \overline{X}^{D}$$. So $$X$$ ⟚ $$\neg \overline{X}^{D}$$ in this case too.

Here the inductive hypothesis is simply that $$Y$$ ⟚ $$\neg \overline{Y}^{D}$$. Hence $$\neg Y$$ ⟚ $$\neg \neg \overline{Y}^{D}$$. But $$\neg \overline{Y}^{D} = \overline{\neg Y}^{D} = \overline{X}^{D}$$. 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 $$\neg P_i$$ for $$P_i$$. Hence, $$\neg \psi \models \neg \phi$$, by contraposition; so finally, $$\psi^D \models \phi^D$$, by the property that $$\varphi^{D}$$ ⟚ $$\neg \overline{\varphi}$$, which was just proved above. And since $$\varphi^{DD} = \phi$$, it is also true that $$\varphi \models \psi$$ if, and only if, $$\psi^D \models \phi^D$$. And it follows, as a corollary, that if $$\phi \models \neg \psi$$, then $$\phi^D \models \neg \psi^D$$.

Conjunctive and disjunctive normal forms
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 $$\neg \varphi$$. This provides a procedure for converting between conjunctive normal form and disjunctive normal form. 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.