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**method 1:**

[(p$\leftrightarrow$q)$\wedge$(q$\leftrightarrow$r)]$\rightarrow$(p$\leftrightarrow$r)

try to make T$\rightarrow$F , So we make RHS false by p is T and r is F. Now, in LHS either q isT Or q is F, LHS always false.

So, we can conclude that above logic statement is a **tautology**.

**method 2:**

[(p$\leftrightarrow$q)$\wedge$(q$\leftrightarrow$r)]$\rightarrow$(p$\leftrightarrow$r)

Convert into boolean expression,

= [($\overline{\bar{p}\bar{q}+pq)(\bar{q}\bar{r}+qr}$)]$+$($\bar{p}\bar{r}+pr$)

= [($p+q$)($\bar{p}+\bar{q}$)]$+$[($q+r$)($\bar{q}+\bar{r}$)]$+$($\bar{p}\bar{r}+pr$)

= $\bar{p}q+p\bar{q}+\bar{q}r+q\bar{r}+\bar{p}\bar{r}+pr$

Now, apply double negation law

= $\overline{\overline{\bar{p}q+p\bar{q}+\bar{q}r+q\bar{r}+\bar{p}\bar{r}+pr}}$

= $\overline{(p+\bar{q})(\bar{p}+q)(q+\bar{r})(\bar{q}+r)(p+r)(\bar{p}+\bar{r})}$

= $\overline{(\bar{p}\bar{q}+pq)(\bar{q}\bar{r}+qr)(\bar{p}r+p\bar{r})}$

= $\overline{(\bar{p}\bar{q}\bar{r}+pqr)(\bar{p}r+p\bar{r})}$

= $\overline{0}$

= $1$ **(tautology) **