Let $X_{i}\,(i=1,2,3,4,5,6)$ be i.i.d. continuous random variables with distribution $F( \cdot )$ and density $f( \cdot )$. What will be the distribution of $$\min(X_1+X_2+X_3,~ X_2+X_3+X_4,~ X_3+X_4+X_5,~ X_4+X_5+X_6)~?$$
Here is my attempt-
Let $W=\min(X_1+X_2+X_3,X_2+X_3+X_4,X_3+X_4+X_5,X_4+X_5+X_6)$. We need to find $P(W\leq w)$. Let $X_2 = x_2,X_3=x_3,X_4=x_4,X_5=x_5$. Then, the problem reduces to
$$P\left( \begin{aligned} &X_1 \leq w-x_2-x_3, \\ &x_2+x_3+x_4\leq w, \\ &x_3+x_4+x_5\leq w, \\ &X_6\leq w-x_4-x_5 \end{aligned}~\middle|~ \begin{aligned} X_2 &= x_2, \\ X_3 &= x_3, \\ X_4 &= x_4, \\ X_5 &=x_5 \end{aligned}\right)~$$
Since all $X_i$ are independent, then we can write the above expression as
$$P(X_1\leq w-x_2-x_3) \cdot P(X_6\leq w-x_4-x_5)$$ subject to the limits $$x_2+x_3+x_4\leq w,x_3+x_4+x_5\leq w~.$$
Is my approach correct?