Question

Show that each of these conditional statements is a tautology by using truth tables.

(a) $\mathbf{[}\mathbf{\neg }\mathbf{p}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{]}\mathbf{\to }\mathbf{q}$

(b) $\mathbf{\left[}\mathbf{\right(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}$

(c) $\mathbf{[}\mathbf{p}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{q}$

(d) $\mathbf{\left[}\mathbf{\right(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{r}$

Short answer

For showing the given conditional statement is a tautology, it is proved that the outcome of these statements in a truth table contains an onlyTvalue.

Step-by-step solution

Definition of Truth Tables  

A logic gatetruth table depicts each feasible input sequence to the gate or circuit, as well as the resulting output based on the combination of these inputs.

To Show conditional statement is a tautology using a truth table

(a) $\mathbf{[}\mathbf{\neg }\mathbf{p}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{q}$

Prepare the truth table for $\mathbf{[}\mathbf{\neg }\mathbf{p}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{q}$.

p	q	localid="1668170251222" $\mathbf{\neg }\mathbf{p}$	localid="1668170264238" $\mathbf{p}\mathbf{\vee }\mathbf{q}$	localid="1668170278223" $\mathbf{\neg }\mathbf{p}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}$	$\mathbf{[}\mathbf{\neg }\mathbf{p}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{q}$
T	T	F	T	F	T
T	F	F	T	F	T
F	T	T	T	T	T
F	F	T	F	F	T

Truth Table

The conditional statement $\mathbf{[}\mathbf{\neg }\mathbf{p}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{q}$is a tautology as the output of the truth table consists of only $\mathbf{T}$.

(b) $\mathbf{\left[}\mathbf{\right(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}$

Prepare the truth table for $\mathbf{\left[}\mathbf{\right(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}$.

p	q	r	$\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}$	$\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{)}$	$\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{)}$	$\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}$	$\mathbf{\left[}\mathbf{\right(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}$
T	T	T	T	T	T	T	T
T	T	F	T	F	F	F	T
T	F	T	F	T	F	T	T
T	F	F	F	T	F	F	T
F	T	T	T	T	T	T	T
F	T	F	T	F	F	T	T
F	F	T	T	T	T	T	T
F	F	F	T	T	T	T	T

Truth Table

The conditional statement localid="1668173717311" $\mathbf{\left[}\mathbf{\right(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}$is a tautology as the output of the truth table consists of only T.

(c) $\mathbf{[}\mathbf{p}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{q}$

Prepare the truth table for $\mathbf{[}\mathbf{p}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{q}$

p	q	$\mathbf{p}\mathbf{\to }\mathbf{q}$	$\mathbf{p}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}$	$\mathbf{\neg }\mathbf{p}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}$
T	T	T	T	T
T	F	F	F	T
F	T	T	F	T
F	F	T	F	T

Truth Table

The conditional statement $\mathbf{[}\mathbf{p}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{q}$is a tautology as the output of the truth table consists of only T.

(d) $\mathbf{\left[}\mathbf{\right(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{r}$

Prepare the truth table for localid="1668174934663" $\mathbf{\left[}\mathbf{\right(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{r}$ .

Suppose$\mathbf{E}\mathbf{=}\mathbf{\left[}\mathbf{\right(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{r}$

p	q	r	$\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}$	$\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}$	$\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{)}$	$\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}$	$\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{)}$	$\mathbf{E}$
T	T	T	T	T	T	T	T	T
T	T	F	T	F	F	F	T	T
T	F	T	F	T	F	T	T	T
T	F	F	F	T	F	F	T	T
F	T	T	T	T	T	T	T	T
F	T	F	T	F	F	T	T	T
F	F	T	T	T	T	T	T	T
F	F	F	T	T	T	T	T	T

Truth Table

The conditional statement $\mathbf{\left[}\mathbf{\right(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{\left)}\mathbf{\right]}\mathbf{\to }\mathbf{r}$is a tautology as the output of the truth table consists of onlyT.

Therefore, it has been shown that the given conditional statements are tautologies using truth tables.