Chapter 3

Chapter 3 - Chapter 3 Problem Solutions-y 3.1(a xy xy x =...

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- 3.1 - Chapter 3 Problem Solutions 3.1. (a) xy + xy - + x - y - = x(y+y - ) + x - y - P4b = x . 1 + x - y - P5a = x + x - y - P2b = x + y - T7a (b) (x - z - + x - y + x - z + xy) __________________ = (x - z + x - z - + yx + yx - ) __________________ P3a,P3b = [x - (z+z - ) + y(x+x - )] ________________ P4b = (x - . 1 + y . 1) __________ P5a = (x - + y) ______ P2b = (x - ) __ y - T9a = xy - T5 (c) (x + y)(x - z - + z)(y - + xz) _______ = (x + y)(z + z - x - )(y - + xz) _______ P3a,P3b = (x + y)(z + x - )(y - + xz) _______ T7a = (x + y)(z + x - )y(xz) ___ T9a,T5 = (x + y)(z + x - )y(x - + z - )T 9 b = (x + y)(x - + z)(x - + z - )y P3a,P3b = (x + y)(x - + zz - )y P4a = (x + y)(x - + 0)y P5b = (x + y)x - yP 2 a = x - y(x + y) P3b = x - [y(x + y)] T8b = x - [y(y + x)] P3a = x - yT 6 b

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- 3.2 - 3.1. (continued) (d) (xy + yz + xz) _____________ = (xy) ___ (yz) ___ (xz) ___ T9a = (x - + y - )(y - + z - )(x - + z - )T 9 b = (x - + y - )(x - + z - )(y - + z - )P 3 b = (x - + y - z - )(y - + z - 4 a = (x - + y - z - )y - + (x - + y - z - )z - P4b = y - (x - + y - z - ) + z - (x - + y - z - 3 b = y - x - + y - y - z - + z - x - + z - y - z - P4b = x - y - + y - y - z - + x - z - + y - z - z - P3b = x - y - + y - z - + x - z - + y - z - T4b = x - y - + y - z - + y - z - + x - z - P3a = x - y - + y - z - + x - z - T4a (e) xy + yz + x - z = xy + yz . 1 + x - zP 2 b = xy + yz(x + x - ) + x - 5 a = xy + yzx + yzx - + x - 4 b = xyz + xy + x - zy + x - z P3a,P3b = xy(z + 1) + x - z(y + 1) P4b = xy . 1 + x - z . 1T 2 a = xy + x - 2 b (f) (x + y)(x - + z) = (x + y)x - + (x + y)z P4b = x - (x + y) + z(x + y) P3b = x - x + x - y + zx + zy P4b = xx - + x - y + zx + zy P3b = 0 + x - y + zx + zy P5b = x - y + zx + zy P2a = x - y + zx + zy . 1P 2 b = x - y + zx + zy(x + x - 5 a = x - y + zx + zyx + zyx - P4b = x - yz + x - y + xzy + xz P3a,P3b = x - y(z + 1) + xz(y + 1) P4b = x - y . 1 + xz . 2 a = x - y + xz P2b
- 3.3 - 3.1. (continued) (g) (x + y)(y + z)(x + z) = [(x + y)y + (x + y)z](x + z) P4b = [y(x + y) + z(x + y)](x + z) P3b = (yx + yy + zx + zy)(x + z) P4b = (yx + y + zx + zy)(x + z) T4b = (y + yx + yz + zx)(x + z) P3a = (y + zx)(x + z) T6a = (y + zx)x + (y + zx)z P4b = x(y + zx) + z(y + zx) P3b = xy + xzx + zy + zzx P4b = xy + xxz + yz + xzz P3b = xy + xz + yz + xz T4b = xy + (xz + xz) + yz P3a = xy + xz + yz T4a (h) xy - + yz - + x - z = xy - . 1 + yz - . 1 + x - z . 1P 2 b = xy - (z + z - ) + yz - (x + x - ) + x - z(y + y - )P 5 a = xy - z + xy - z - + yz - x + yz - x - + x - zy + x - zy - P4b = x - yz + x - yz - + y - zx + y - zx - + xz - y + xz - y - P3a,P3b = x - y(z + z - ) + y - z(x + x - ) + xz - (y + y - 4 b = x - y . 1 + y - z . 1 + xz - . 5 a = x - y + y - z + xz - P2b 3.2. To prove the cancellation law does not hold, use the method of contradiction. According to T7b, x(x - +y)=xy. Assuming the cancellation law holds, it follows that x - +y=y for all x and y in the Boolean algebra. Since x and y denote any elements in the Boolean algebra, let x=y. It then follows that y - +y=y. However, from P5a it is known that y must be 1 in this case and not any arbitrary element in the algebra. Thus, by contradiction, the cancellation law does not hold. By applying a dual argument starting with T7a, it can be shown that x+y=x+z does not imply y=z.

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- 3.4 - 3.3. (a) (b) (c) xyzx - z+xy (x - +y)(x+z) (x+y+z) ______ x - y - z - xy+yz+x - zx y + x - z 000 o 0 1 1 0 0 001 1 1 0 0 1 1 010 0 0 0 0 0 0 011 1 1 0 0 1 1 100 0 0 0 0 0 0 101 0 0 0 0 0 0 110 1 1 0 0 1 1 111 1 1 0 0 1 1 3.4. Let B={0,1,a} where a 0,1. By Postulate P5 of a Boolean algebra, the complement of the element a, i.e., a - , must exist and satisfy the relationships a+a - =1 and a . a - =0. Since there are just three elements in B, a - must be 0, 1, or a. Suppose a - =a, then a+a - =a+a=a. However, since a 1, Postulate P5a is not satisfied. Thus, a - a. Now suppose a - =1. Then a .
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Chapter 3 - Chapter 3 Problem Solutions-y 3.1(a xy xy x =...

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