Question

Show that if p,q and r are compound propositions such that p and qare logically equivalent and q and rare logically equivalent, then pand r are logically equivalent.

Short answer

For finding the number of truth tables of compound propositions that contains the propositional variables pand q , find the number of input variables and the rows in the truth table.

Step-by-step solution

Definition of logically equivalent propositions

Any two propositions are said to be logically equivalent if they have same truth value for any combination of truth values of p and q .

To show p and r are logically equivalent.

Take three input variables such asp,qandr.

Ifpandqare logically equivalent, then $\mathbf{p}\mathbf{\leftrightarrow }\mathbf{q}$is a tautology.

Similarly, ifqandrare logically equivalent, then $\mathbf{q}\mathbf{\leftrightarrow }\mathbf{r}$is a tautology.

Also, the conjunction of two tautologies is also a tautology.

That means, $\left(\mathbf{p}\mathbf{\leftrightarrow }\mathbf{q}\right)\mathbf{\wedge }\left(q\leftrightarrow r\right)$is a tautology.

With respect to exercise 29, $\left(\mathbf{p}\mathbf{\to }\mathbf{q}\right)\mathbf{\wedge }\left(q\to r\right)\mathbf{\to }\left(p\to r\right)$

Thus, $\left(\mathbf{p}\mathbf{\leftrightarrow }\mathbf{q}\right)\mathbf{\wedge }\mathbf{\left(}\mathbf{q}\mathbf{\leftrightarrow }\mathbf{r}\mathbf{\right)}$is equivalent with $\left(\mathbf{p}\mathbf{\to }\mathbf{r}\right)$.

Follow logical equivalence law.

$\left(\mathbf{p}\mathbf{\leftrightarrow }\mathbf{q}\right)\mathbf{\wedge }\left(q\leftrightarrow r\right)\mathbf{\equiv }\left(\left(p\to q\right)\wedge \left(q\to p\right)\right)\mathbf{\wedge }\left(\left(q\to r\right)\wedge \left(r\to p\right)\right)$

Now, use commutative and associative law

$\left(\mathbf{p}\mathbf{\leftrightarrow }\mathbf{q}\right)\mathbf{\wedge }\mathbf{\left(}\mathbf{q}\mathbf{\leftrightarrow }\mathbf{r}\mathbf{\right)}\mathbf{\equiv }\mathbf{(}\mathbf{\left(}\mathbf{p}\mathbf{\to }\mathbf{q}\mathbf{\right)}\mathbf{\wedge }\mathbf{(}\mathbf{q}\mathbf{\to }\mathbf{r}\mathbf{)}\mathbf{)}\mathbf{\wedge }\mathbf{(}\mathbf{\left(}\mathbf{r}\mathbf{\to }\mathbf{q}\mathbf{\right)}\mathbf{\wedge }\mathbf{\left(}\mathbf{q}\mathbf{\to }\mathbf{p}\mathbf{\right)}\mathbf{)}$

Use the result of exercise 29

$\left(\mathbf{p}\mathbf{\leftrightarrow }\mathbf{q}\right)\mathbf{\wedge }\left(q\leftrightarrow r\right)\mathbf{\equiv }\left(\left(p\to r\right)\wedge \left(r\to p\right)\right)$

Follow the logical equivalenace law

$\left(\mathbf{p}\mathbf{\leftrightarrow }\mathbf{q}\right)\mathbf{\wedge }\mathbf{\left(}\mathbf{q}\mathbf{\leftrightarrow }\mathbf{r}\mathbf{\right)}\mathbf{\equiv }\mathbf{p}\mathbf{\leftrightarrow }\mathbf{r}$

Therefore,pis logically equivalent torif $\mathbf{p}\mathbf{\leftrightarrow }\mathbf{r}$is logically equivalent with $\left(\mathbf{p}\mathbf{\leftrightarrow }\mathbf{q}\right)\mathbf{\wedge }\mathbf{\left(}\mathbf{q}\mathbf{\leftrightarrow }\mathbf{r}\mathbf{\right)}\mathbf{\equiv }\mathbf{p}\mathbf{\leftrightarrow }\mathbf{r}$and $\mathbf{p}\mathbf{\leftrightarrow }\mathbf{r}$is a tautology.