Language $L_1$ is polynomial time reducible to language $L_2$. Language $L_3$ is polynomial time reducible to $L_2$, which in turn is polynomial time reducible to language $L_4$. Which of the following is/are true? I. if $L_4 \in P$, then $L_2 \in P$ II. if $L_1 \in P$ or $L_3 \in P$, then $L_2 \in P$ III. $L_1 \in P$, if and only if $L_3 \in P$ IV. if $L_4 \in P$, then $L_1 \in P$ and $L_3 \in P$

We are given the following polynomial-time reductions:

• $L_1 \le_p L_2$

• $L_3 \le_p L_2$

• $L_2 \le_p L_4$

By the transitivity of polynomial-time reductions:

• Since $L_1 \le_p L_2$ and $L_2 \le_p L_4$, we have $L_1 \le_p L_4$.

• Since $L_3 \le_p L_2$ and $L_2 \le_p L_4$, we have $L_3 \le_p L_4$.

A fundamental property of polynomial-time reducibility ($A \le_p B$) is that if $B \in \text{P}$, then $A \in \text{P}$.

Now, evaluate each statement:

• Statement I: If $L_4 \in \text{P}$, since $L_2 \le_p L_4$, it follows that $L_2 \in \text{P}$. Thus, statement I is true.

• Statement II: If $L_1 \in \text{P}$ or $L_3 \in \text{P}$, it does not imply $L_2 \in \text{P}$, because a hard problem can reduce to an easier one, but not necessarily vice versa. Thus, statement II is false.

• Statement III: $L_1$ and $L_3$ both reduce to $L_2$, but they are not necessarily reducible to each other. One can be in $\text{P}$ while the other is not. Thus, statement III is false.

• Statement IV: If $L_4 \in \text{P}$, then since $L_1 \le_p L_4$ and $L_3 \le_p L_4$, both $L_1 \in \text{P}$ and $L_3 \in \text{P}$. Thus, statement IV is true.

Therefore, statements I and IV only are true, corresponding to option (C).