The product of maxterms for the function F(A, B, C, D) = D(A′ + B) + B′D is

Concept. In a canonical product-of-maxterms form ΠM(…), the listed indices are exactly the input rows on which the function evaluates to 0; in a canonical sum-of-minterms form Σm(…), the listed indices are the rows on which it evaluates to 1. For n variables the 2n row indices are split between the two lists, so an index is a maxterm precisely when it is not a minterm. The complement law B + B′ = 1 together with X + 1 = 1 is what collapses a redundant expression before its rows are read off.

Application.

1. Expand the expression: F = D(A′ + B) + B′D = A′D + BD + B′D.

2. Combine the two terms that differ only in B: BD + B′D = D(B + B′) = D · 1 = D.

3. Substitute back: F = A′D + D = D(A′ + 1) = D · 1 = D. The function therefore depends on D alone; A, B and C do not affect its value.

4. Read off the rows. With the standard weights A = 8, B = 4, C = 2, D = 1, the row index is 8A + 4B + 2C + D, so D = 1 holds exactly on the odd indices 1, 3, 5, 7, 9, 11, 13, 15 and D = 0 holds exactly on the even indices 0, 2, 4, 6, 8, 10, 12, 14.

5. A product of maxterms collects the rows on which F = 0, which here are the even indices: F = ΠM(0, 2, 4, 6, 8, 10, 12, 14). The complementary canonical form is F = Σm(1, 3, 5, 7, 9, 11, 13, 15).

Cross-check.

• Row 10 (A = 1, B = 0, C = 1, D = 0): F = 0 · (0 + 0) + 1 · 0 = 0, so index 10 is a zero row and belongs in the maxterm list.

• Row 7 (A = 0, B = 1, C = 1, D = 1): F = 1 · (1 + 1) + 0 · 1 = 1, so index 7 is a one row and stays out of the maxterm list.

• Count: 8 maxterm indices + 8 minterm indices = 16 = 24 rows, so the two lists partition the truth table with no index repeated or missing.

• Because C never appears in F, two rows that differ only in C must fall in the same list — 0 with 2, 4 with 6, 8 with 10, 12 with 14 — which the even-index list satisfies.

Result. F(A, B, C, D) = D, so the product of maxterms is ΠM(0, 2, 4, 6, 8, 10, 12, 14).