Given the function $F = P^{\prime} + QR$, where $F$ is a function in three Boolean variables $P, Q$ and $R$ and $P^{\prime} = !P$, consider the following statements. (S1) $F = \sum (4, 5, 6)$ (S2) $F = \sum (0, 1, 2, 3, 7)$ (S3) $F = \prod (4, 5, 6)$ (S4) $F = \prod (0, 1, 2, 3, 7)$ Which of the following is true?

To express the Boolean function $F(P, Q, R) = P^{\prime} + QR$ in canonical sum-of-minterms (SOP) and product-of-maxterms (POS) forms:

1. Expand the terms into canonical minterms:

   $$P^{\prime} = P^{\prime}(Q + Q^{\prime})(R + R^{\prime}) = P^{\prime}Q^{\prime}R^{\prime} + P^{\prime}Q^{\prime}R + P^{\prime}QR^{\prime} + P^{\prime}QR = m_0 + m_1 + m_2 + m_3$$

   $$QR = (P + P^{\prime})QR = PQR + P^{\prime}QR = m_7 + m_3$$

2. Combining these gives the minterm expansion:

   $$F(P, Q, R) = m_0 + m_1 + m_2 + m_3 + m_7 = \sum (0, 1, 2, 3, 7)$$

   Thus, statement (S2) is True and (S1) is False.

3. The maxterms of $F$ correspond to the minterms not present in $F$, which are $4, 5, 6$:

   $$F(P, Q, R) = M_4 \cdot M_5 \cdot M_6 = \prod (4, 5, 6)$$

   Thus, statement (S3) is True and (S4) is False.

Hence, the correct evaluation of the statements is: (S1)- False, (S2)- True, (S3)- True, (S4)- False, which corresponds to option (A).