Design a K-Map for \(F(x, y, z) = x\overline{z} + xyz + y\overline{z}\) of \(F(x, y, z)\) and find prime implicants.
\(xy, y\overline{z}, x\overline{z}\)
Write each product term as minterms (order x, y, z): \(x\overline{z}\) covers 100 and 110, i.e. \(m_{4}, m_{6}\); \(xyz\) is 111, i.e. \(m_{7}\); \(y\overline{z}\) covers 010 and 110, i.e. \(m_{2}, m_{6}\).
So \(F = \sum m(2, 4, 6, 7)\). On the 3-variable K-map these four 1-cells are placed and grouped into the largest possible rectangles of size 2 (no group of 4 exists, because \(m_{0}\), \(m_{3}\) and \(m_{5}\) are 0).
The pair \(m_{6}, m_{7}\) (110, 111) gives \(xy\); the pair \(m_{2}, m_{6}\) (010, 110) gives \(y\overline{z}\); the pair \(m_{4}, m_{6}\) (100, 110) gives \(x\overline{z}\). None of these can be enlarged, so all three are prime implicants.
Terms such as \(yz\), \(\overline{x}y\) or \(\overline{x}z\) would include the 0-cells \(m_{3}\) or \(m_{1}\), and single literals x, y, z each cover cells where F = 0, so they cannot be implicants.
Check: \(xy + y\overline{z} + x\overline{z}\) is 1 exactly for 010, 100, 110 and 111, which reproduces \(\sum m(2, 4, 6, 7)\).
Hence, \(xy, y\overline{z}, x\overline{z}\).
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