Question

Show that each conditional statement in the Exercise 10 is a tautology without using truth tables.

Short answer

For showing the given conditional statements are tautologies, it is proved that the outcome of these statements in a truth table contains only Tvalue.

Step-by-step solution

Definition of Tautology  

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

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

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

Follow the rule of logical equivalence

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

Follow De Morgan’s law

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

Follow double negation law

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

Follow De Morgan’s law

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

Follow distributive law

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

Follow negation law

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

Follow commutative law

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

Follow identity law

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

Follow associative and commutative law

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

Follow negation law

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

Follow domination law

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

Therefore, the conditional statement $\mathbf{[}\mathbf{\neg }\mathbf{p}\mathbf{\wedge }\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}\mathbf{]}\mathbf{\to }\mathbf{q}$ is a tautology as it equivalent to

T

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

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

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

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

Follow the rule of logical equivalence

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

Follow De Morgan’s law twice and the double negation law

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

Follow distributive law twice

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

Follow negation law and commutative law

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

Follow identity law

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

Follow commutative and associative law

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

Follow commutative law and domination law

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

Follow identity law

role="math" localid="1668241157691" $\mathbf{\left[}\left(p\to q\right)\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{)}\mathbf{\right]}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}\mathbf{=}\mathbf{\left[}\left((p\vee q)\wedge (\neg q\vee \neg r))\vee (\neg p\vee r\right)\mathbf{\right]}$

Follow distributive law

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

Follow Commutative law and associative law

role="math" localid="1668241646783" $\mathbf{\left[}\left(p\to q\right)\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{)}\mathbf{\right]}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}\mathbf{=}\mathbf{\left[}\left((p\vee \neg q)\wedge (q\vee r)\right)\mathbf{\vee }\mathbf{\left(}\mathbf{\right(}\mathbf{\neg }\mathbf{q}\mathbf{\vee }\mathbf{\neg }\mathbf{p}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{\neg }\mathbf{r}\mathbf{\vee }\mathbf{r}\mathbf{\left)}\mathbf{\right)}\mathbf{\right]}$

Follow negation law and commutative law

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

Follow domination law

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

Follow identity law

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

The conditional statement $\left[\left(p\to q\right)\wedge (q\to r)\right]\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{r}\mathbf{)}$ is a tautology as it equivalent to

T.

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

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

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

role="math" localid="1668242496181" $\mathbf{\left[}\mathbf{p}\mathbf{\wedge }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{\right]}\mathbf{\to }\mathbf{q}\mathbf{=}\mathbf{T}$

Follow the rule of logical equivalence

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

Follow De Morgan’s law

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

Follow negation law

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

Follow identity law

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

Follow the rule of logical equivalence

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

Follow negation and commutative law

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

Follow domination law

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

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

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

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

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

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

Follow the rule of logical equivalence

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

Follow distributive law

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

Follow the rule of logical equivalence

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

Follow De Morgan’s law thrice and the double negation law

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

Follow distributive law

role="math" localid="1668248271303" $\mathbf{(}\mathbf{p}\mathbf{\wedge }\mathbf{q}\mathbf{)}\mathbf{\to }\mathbf{(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{)}\mathbf{=}\mathbf{\left[}\mathbf{\right(}\mathbf{(}\mathbf{\neg }\mathbf{p}\mathbf{\vee }\mathbf{\neg }\mathbf{q}\mathbf{)}\mathbf{\vee }\mathbf{(}\mathbf{p}\mathbf{\vee }\mathbf{q}\mathbf{)}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{(}\mathbf{\neg }\mathbf{p}\mathbf{\wedge }\mathbf{\neg }\mathbf{q}\mathbf{)}\mathbf{\vee }\mathbf{\neg }\mathbf{r}\mathbf{)}\mathbf{]}\mathbf{\vee }\mathbf{r}$

Follow commutative and associative law

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

Follow negation law

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

Follow domination law

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

Follow identity law

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

Follow distributive law

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

Follow negation and commutative law

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

Follow domination law

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

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 equivalent to T

Therefore, all the given conditional statements are tautologies.