Question

Explain, without using a truth table, why \((p \vee \neg q) \wedge (q \vee \neg r) \wedge (r \vee \neg p)\) is true when \(p,\;q\), and \(r\) have the same truth value and it is false otherwise.

Short answer

$(\mathrm{p}\vee \neg \mathrm{q})\wedge (\mathrm{q}\vee \neg \mathrm{r})\wedge (\mathrm{r}\vee \neg \mathrm{p})$ is true when p, q & r have same truth value and it is false otherwise.

Step-by-step solution

Definition of truth table

A truth table is a mathematical table which is used in logic.

Prove the given statement

Let p, q & r are true,

$(\mathrm{p}\vee \neg \mathrm{q})$is true.

Similarly $(\mathrm{q}\vee \neg \mathrm{r})\&(\mathrm{r}\vee \neg \mathrm{p})$are true.

The statement becomes true.

Similarly, when p, q & r are false then statement $(\mathrm{p}\vee \neg \mathrm{q})\wedge (\mathrm{q}\vee \neg \mathrm{r})\wedge (\mathrm{r}\vee \neg \mathrm{p})$becomes true.

When p,q are true and ris falsethen $(\mathrm{p}\vee \neg \mathrm{q})$ is true$(\mathrm{q}\vee \neg \mathrm{q})$is truebut$(\mathrm{r}\vee \neg \mathrm{p})$is false.

The statement $(\mathrm{p}\vee \neg \mathrm{q})\wedge (\mathrm{q}\vee \neg \mathrm{r})\wedge (\mathrm{r}\vee \neg \mathrm{p})$becomes false.

Hence, we proved $(\mathrm{p}\vee \neg \mathrm{q})\wedge (\mathrm{q}\vee \neg \mathrm{r})\wedge (\mathrm{r}\vee \neg \mathrm{p})$is true when p,q,r have same truth value and it is false otherwise.