Question

Why are the duals of two equivalent compound propositions also equivalent, where these compound propositions contain only the operators$\wedge ,\vee ,$ and $\neg$?

Short answer

Duals of equivalent compound compositions are also equivalent.

Step-by-step solution

LOGICAL EQUIVALENCES

LOGICAL EQUIVALENCES

Identity laws:

$$\begin{gathered} p\vee F\equiv p \\ p\wedge T\equiv p \end{gathered}$$

DOMINATION LAWS:

$$\begin{gathered} p\wedge F\equiv F \\ p\vee T\equiv T \end{gathered}$$

IDEMPOTENT LAWS:

$$\begin{gathered} p\wedge p\equiv p \\ p\vee p\equiv p \end{gathered}$$

COMMUTATIVE LAWS:

$$\begin{gathered} p\wedge q\equiv q\wedge p \\ p\vee q\equiv q\vee p \end{gathered}$$

ASSOCIATIVE LAWS:

$$\begin{gathered} (p\wedge q)\wedge r\equiv p\wedge (q\wedge r) \\ (p\vee q)\vee r\equiv p\vee (q\vee r) \end{gathered}$$

DISTRIBUTIVE LAWS

$$\begin{gathered} p\wedge (q\vee r)\equiv (p\wedge q)\vee (p\wedge r) \\ p\vee (q\wedge r)\equiv (p\vee q)\wedge (p\vee r) \end{gathered}$$

DEMORGAN'S LAWS

$$\begin{gathered} \neg (p\vee q)\equiv \neg p\wedge \neg q \\ \neg (p\wedge q)\equiv \neg p\vee \neg q \end{gathered}$$

ABSORPTION LAW

$$\begin{gathered} p\wedge (p\vee q)\equiv p \\ p\vee (p\wedge q)\equiv p \end{gathered}$$

NEGATION LAWS

$$\begin{gathered} p\wedge \neg p\equiv F \\ p\vee \neg p\equiv T \end{gathered}$$

Solution

In the previous exercise, we derived that the dual of every logical equivalence in the above laws is the other logical equivalence in the same law. Moreover, the dual of the double negation law is the double negation law itself. This then means that the dual of every logical equivalence law is also a logical equivalence law and thus the duals of equivalent compound compositions are also equivalent.

Identity Laws:

The dual replaces$\wedge$by V , Vby$\wedge ,T$by F and F by T.

The dual of $p\wedge T\equiv p$is then$p\vee F\equiv p$

The dual of$p\vee F\equiv p$ is then$p\wedge T\equiv p$

Domination Laws:

The dual replaces$\wedge$ by V,Vby $\wedge ,T$by F and F by T.

The dual of$p\vee T\equiv T$ is then$p\wedge F\equiv F$

The dual of $p\wedge F\equiv F$is then$p\vee T\equiv T$

IDEMPOTENT LAWS

The dual replaces$\wedge$ by$\vee ,\vee$by$\wedge ,T$byF and Fby T.

The dual of$p\vee p\equiv p$is then$p\wedge p\equiv p$

The dual of$p\wedge p\equiv p$ is then$p\vee p\equiv p$

COMMUTATIVE LAWS

The dual replaces $\wedge$by$\vee ,\vee$ by$\wedge ,T$byF and F byT .

The dual of $p\vee q\equiv q\vee p$is then $p\wedge q\equiv q\wedge p$

The dual of $p\wedge q\equiv q\wedge p$is then$p\vee q\equiv q\vee p$

ASSOCIATIVE LAWS

The dual replaces$\wedge$ by$\vee ,\vee$ by$\wedge ,T$ byF and F byT .

The dual of $(p\vee q)\vee r\equiv p\vee (q\vee r)$is then $(p\wedge q)\wedge r\equiv p\wedge (q\wedge r)$

The dual of$(p\wedge q)\wedge r\equiv p\wedge (q\wedge r)$ is then$(p\vee q)\vee r\equiv p\vee (q\vee r)$

DISTRIBUTIVE LAWS

The dual replaces$\wedge$ by $\vee ,\vee$by$\wedge ,T$ byF andF byT.

The dual of $p\vee (q\wedge r)\equiv (p\vee q)\wedge (p\vee r)$is then $p\wedge (q\vee r)\equiv (p\wedge q)\vee (p\wedge r)$

The dual of $p\wedge (q\vee r)\equiv (p\wedge q)\vee (p\wedge r)$is then $p\vee (q\wedge r)\equiv (p\vee q)\wedge (p\vee r)$

DEMORGAN’S LAWS

The dual replaces$\wedge$ byV,V$\wedge ,T$by T byF and F by T.

The dual of$\neg (p\wedge q)\equiv \neg p\vee \neg q$is then$\neg (p\vee q)\equiv \neg p\wedge \neg q$

The dual of$\neg (p\vee q)\equiv \neg p\wedge \neg q$ is then$\neg (p\wedge q)\equiv \neg p\vee \neg q$

ABSORPTION LAW

The dual replaces ^by byV,V by ^,Tand F by T.

The dual of$p\vee (p\wedge q)\equiv p$is then$p\wedge (p\vee q)\equiv p$

The dual of $p\wedge (p\vee q)\equiv p$is then$p\vee (p\wedge q)\equiv p$

NEGATION LAWS

The dual replaces$\wedge$ byV,V by $\wedge ,T$by F and F by T.

The dual of$p\vee \neg p\equiv T$is then$p\wedge \neg p\equiv F$

The dual of$p\wedge \neg p\equiv F$is then$p\vee \neg p\equiv T$

SOLUTION

In the previous exercise (and also above this solution), we derived that the dual of every logical equivalence in the above laws is the other logical equivalence in the same law. Moreover, the dual of the double negation law is the double negation law itself. This then means that the dual of every logical equivalence law is also a logical equivalence law and thus the duals of equivalent compound compositions are also equivalent.