Question

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

(a)$\mathbf{(}\mathbf{p}\mathbf{\wedge }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{p}$ (b) $\mathbf{p}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}$

(c) role="math" localid="1668154499853" $\mathbf{\neg }\mathbf{p}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}$ (d)$\mathbf{(}\mathit{p}\mathbf{\wedge }\mathit{q}\mathbf{)}\mathbf{\to }\mathbf{(}\mathit{p}\mathbf{\to }\mathit{q}\mathbf{)}$

(e) $\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{p}$ (f)$\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{\neg }\mathbf{q}$

Short answer

It is shown that the conditional statement $\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{\neg }\mathbf{q}$is a tautology as the output of the truth table consists of only T.

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)$\mathbf{(}\mathbf{p}\mathbf{\wedge }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{p}$

Prepare the truth table for $\mathbf{(}\mathbf{p}\mathbf{\wedge }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{p}\mathbf{.}$

p	q	localid="1668156225147" $\mathbf{p}\mathbf{\wedge }\mathbf{q}$	$\mathbf{(}\mathbf{p}\mathbf{\wedge }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{p}$
T	T	T	T
T	F	F	T
F	T	F	T
F	F	F	T

Truth Table

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

(b) $\mathbf{p}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}$

Prepare the truth table for $\mathbf{p}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}$.

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

Truth Table

The conditional statement is a tautology as the output of the truth table consists of onlyT.

(c) $\mathbf{\neg }\mathbf{p}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}$

Prepare the truth table for $\mathbf{\neg }\mathbf{p}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}$

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

Truth Table

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

(d) $\mathbf{(}\mathbf{p}\mathbf{\wedge }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}$

Prepare the truth table for localid="1668162884798" $\mathbf{(}\mathbf{p}\mathbf{\wedge }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{.}$

p	q	$\mathbf{p}\mathbf{\wedge }\mathbf{q}$	$\mathbf{p}\mathbf{\to }\mathbf{q}$	localid="1668163037369" $\mathbf{(}\mathbf{p}\mathbf{\wedge }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}$
T	T	T	T	T
T	F	F	F	T
F	T	F	T	T
F	F	F	T	T

Truth Table

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

(e) $\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{p}$

Prepare the truth table for$\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{p}$.

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

Truth Table

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

(f) $\neg \mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{\neg }\mathbf{q}$

Prepare the truth table for $\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{\neg }\mathbf{q}$.

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

Truth Table

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

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