Question

Show that the logical equivalences in Table 6, except for the double negation law, come in pairs, where each pair contains compound propositions that are duals of each other.

Short answer

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

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

Step-by-step solution

IDENTITY LAWS

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

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

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 F by 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 $\wedge ,T$by F and F by T .

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 by T.

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 by T .

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$\wedge$ by$\vee ,\vee$ by$\wedge ,T$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 thendata-custom-editor="chemistry" $\neg (p\wedge q)\equiv \neg p\vee \neg q$

ABSORPTION LAW

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

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 $p\vee \neg p\equiv T$