ΠM(0,2,6,7) using 4:1 MUX

First convert the maxterm list to minterms. A 3-variable function has minterms 0–7; the ones not listed as maxterms are the minterms:

$$Y = \Pi M(0,2,6,7) = \Sigma m(1,3,4,5)$$

Implementation table

Use $A, B$ as select lines ($S_1 = A$, $S_0 = B$) and express each data input in terms of $C$. Input $I_j$ covers minterms $2j$ ($C = 0$) and $2j + 1$ ($C = 1$).

• $I_0$: only minterm 1 present, so $I_0 = C$.
• $I_1$: only minterm 3, so $I_1 = C$.
• $I_2$: both 4 and 5, so $I_2 = 1$.
• $I_3$: neither 6 nor 7, so $I_3 = 0$.

Circuit

             +--------+
  C -------->| I0     |
  C -------->| I1 4:1 |----> Y
  1 (Vcc) -->| I2 MUX |
  0 (GND) -->| I3     |
             +--------+
                S1  S0
                A   B

No extra gate is needed here (no $C'$ input appears).

Check

The MUX equation gives

$$Y = A'B'C + A'BC + AB'(1) + AB(0) = A'C + AB'$$

Y is 0 exactly at 0, 2, 6, 7, which matches $\Pi M(0,2,6,7)$.