Algebraic Manipulation of Boolean Expressions - Answers

1. Z = f(A,B,C) = + B + AB + AC
   = (B + ) + A(C + B)
   = (B + ) + A(C + B)      from T10b
   = B + + AC + AB
   = B( + A) + + AC      from T9a and T8b
   = B + + AC

   Z = f(A,B,C,D) = B + B + BC + A

   = B + B(C + ) + (B + A)      using B (twice) T4a
   = B + B + (B + A)      from T9a, T8b and T10b
   = B(1 + ) + B + A      from T8a
   = B + B + A      from T8a
   = B(1 + + A
   = B + A      from T8a

   Z = f(A,B,C,D) = AB + ABD + AD + ABCD + ACD + ABC + AC

   = AB + ABC + AC + AD      from T9a and T8b

   = AB + AC + AD      from T9a and T8b

   = A(B + D) + AC

   = A(B + D) + AC      from T10a

   = AB + AD + AC