Question

Show that each conditional statement in the Exercise 9 is a tautology without using truth tables.

Short answer

For showing the given conditional statement is a tautology, prove that the outcome of these statements in a truth table contains an only T value.

Step-by-step solution

Definition of Tautology  

A tautology is a compound statement that is true for all feasible truth values of the statements.

To Show conditional statement is a tautology without using a truth table

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

For showing, a given conditional statement is a tautology, prove that

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

Follow the rule of logical equivalence

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

Follow the rule of logical equivalence

role="math" localid="1668249638369" $\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{p}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{p}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{\neg }\mathbf{p}\mathbf{\vee }\mathbf{p}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{p}\mathbf{)}$

Follow commutative law

$\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{p}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{p}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{\neg }\mathbf{p}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{p}\mathbf{)}$

Follow negation law

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

Follow commutative law

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

Follow domination law

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

Therefore, the conditional statement $\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{p}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{p}\mathbf{)}$ is a tautology as it is equivalent to T.

To Show conditional statement is a tautology without using a truth table

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

For showing, a given conditional statement is a tautology, prove that$\mathbf{P}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}\mathbf{=}\mathbf{T}$

Follow the rule of logical equivalence

$\mathbf{P}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}$

Follow the rule of logical equivalence

$\mathbf{P}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{\neg }\mathbf{p}\mathbf{\vee }\mathbf{p}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}$

Follow commutative law

$\mathbf{P}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{\neg }\mathbf{p}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}$

Follow the rule of negation

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

Follow commutative law

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

Follow domination law

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

The conditional statem $\mathbf{P}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}$ is a tautology as it is equivalent to T.

To Show conditional statement is a tautology without using a truth table

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

For showing, a given conditional statement is a tautology, prove that

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

Follow the rule of logical equivalence

$\mathbf{\neg }\mathbf{p}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{=}\mathbf{p}\mathbf{\vee }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}$

Follow the rule of logical equivalence

$\mathbf{\neg }\mathbf{p}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{=}\mathbf{p}\mathbf{\vee }\mathbf{(}\mathbf{\neg }\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}$

Follow associative law

$\mathbf{\neg }\mathbf{p}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{\neg }\mathbf{p}\mathbf{)}\mathbf{\vee }\mathbf{q}$

Follow the rule of negation

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

Follow domination law

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

The conditional statement $\mathbf{\neg }\mathbf{p}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}$ is a tautology as it is equivalent to T.

To Show conditional statement is a tautology without using a truth table

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

For showing, a given conditional statement is a tautology, prove that

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

Follow the rule of logical equivalence

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

Follow the rule of logical equivalence

role="math" localid="1668251413747" $\mathbf{(}\mathbf{p}\mathbf{\wedge }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{=}\mathbf{[}\mathbf{\neg }\mathbf{p}\mathbf{\vee }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{]}\mathbf{\vee }\mathbf{[}\mathbf{\neg }\mathbf{q}\mathbf{\vee }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{]}$

Follow the rule of logical equivalence

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

Follow commutative law and associative law

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

Follow the rule of negation

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

Follow commutative law

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

Follow domination law

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

The conditional statement $\mathbf{(}\mathbf{p}\mathbf{\wedge }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}$ is a tautology as it is equivalent to T.

To Show conditional statement is a tautology without using a truth table

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

For showing, a given conditional statement is a tautology, prove that$\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{p}$

Follow the rule of logical equivalence

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

Follow the rule of logical equivalence

$\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{p}\mathbf{=}\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{\neg }\mathbf{q}\mathbf{\to }\mathbf{p}\mathbf{)}$

Follow the rule of logical equivalence

$\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{p}\mathbf{=}\mathbf{(}\mathbf{\neg }\mathbf{p}\mathbf{\vee }\mathbf{p}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{\neg }\mathbf{q}\mathbf{\to }\mathbf{p}\mathbf{)}$

Follow commutative law

$\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{p}\mathbf{=}\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{\neg }\mathbf{p}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{\neg }\mathbf{q}\mathbf{\to }\mathbf{p}\mathbf{)}$

Follow the rule of negation

role="math" localid="1668255672220" $\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{p}\mathbf{=}\mathbf{T}\mathbf{\vee }\mathbf{(}\mathbf{\neg }\mathbf{p}\mathbf{\to }\mathbf{p}\mathbf{)}$

Follow commutative law

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

Follow domination law

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

The conditional statement $\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{p}$ is a tautology as it is equivalent to T.

To Show conditional statement is a tautology without using a truth table

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

For showing, a given conditional statement is a tautology, prove that

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

Follow the rule of logical equivalence

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

Follow the rule of logical equivalence

$\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{\neg }\mathbf{q}\mathbf{=}\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{\neg }\mathbf{q}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{\neg }\mathbf{q}\mathbf{\to }\mathbf{\neg }\mathbf{q}\mathbf{)}$

Follow the rule of logical equivalence

$\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{\neg }\mathbf{q}\mathbf{=}\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{\neg }\mathbf{q}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{\neg }\mathbf{(}\mathbf{\neg }\mathbf{q}\mathbf{\to }\mathbf{\neg }\mathbf{q}\mathbf{)}$

Follow the double negation law

$\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{\neg }\mathbf{q}\mathbf{=}\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{\neg }\mathbf{q}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{q}\mathbf{\vee }\mathbf{\neg }\mathbf{q}\mathbf{)}$

Follow the rule of negation

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

Follow domination law

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

The conditional statement$\mathbf{\neg }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{\neg }\mathbf{q}$ is a tautology as it is equivalent to T.

Therefore, all the given conditional statements are tautologies.