Solution

The correct answer is Option 4)  xyz + xyz' + xy'z'

Key Points

• Given Boolean expression: A(x,y,z) = x (y' z)'.
• Use De Morgan’s law to remove the outer complement:
  (y' z)' = (y')' + z' = y + z'
• Therefore,
  A = x (y + z') = xy + xz'
  This is a valid SOP, but not complete because each product term must list all variables (minterms).

Make it a complete sum-of-products (expand to minterms)

• Term 1: xy (z is don’t-care)
  xy = xy(z + z') = xyz + xyz'
• Term 2: xz' (y is don’t-care)
  xz' = xz'(y + y') = xyz' + xy'z'
• Combine and remove duplicates:
  A = (xyz + xyz') + (xyz' + xy'z') = xyz + xyz' + xy'z'

Important Points

• Complete SOP (canonical SOP) means a sum of minterms: each product contains all variables in either complemented or uncomplemented form.
• De Morgan’s Law: (AB)' = A' + B' and (A + B)' = A'B'.
• To expand a product with a don’t-care variable, use X = X(Y + Y') to obtain both minterms.