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

  2. 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

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Composed by David Belton - April 98