Week 5 Individual assignment
James Mullen
MTH/221
10/19/13
John Beris

Chapter 15 supplementary question 5

5. Let be a Boolean algebra that is partially ordered by≤. If x, y, z ∈ B, prove that x + y ≤ z if and only if x ≤ z and y ≤ z.

If x ≤ z and y ≤ z, then from Exercise 6(b) of Section 15.4 we have x + y ≤ z + z. And by the Idempotent Law we have z + z = z. Conversely, suppose that x + y ≤ z.We find that x ≤ x + y, because x(x + y) = x + xy (by the Idempotent Law) _ x (by the Absorption Law).
Since x ≤ x + y and x + y ≤ z, we have x ≤ z, because a partial order is transitive. (The proof that y ≤ z follows in a similar way.)

15.1 Question 2
2. Let w, x, and y be Boolean variables where the value of x is 1. For each of the following Boolean expressions, determine, if possible, the value of the expression. If you cannot determine the value of the expression, then find the number of assignments of values for w and y that will result in the value 1 for the expression.
a) x + xy + w b) xy + w
c) xy + xw d) xy + w
a)
if w, x, and y be Boolean variables, where the value of x is 1, we need to determine the value of the expression x + xy + w and if it is not possible to find the value of the expression, then find the number of assignments of value for w and y that will result in the value 1 for the expression.
Using the law of Identity in the given expression: we get x-1 + xy + w = x (1 + y) + w.
Using the law of Dominance, we get x + xy+ w.= x -1+ w.
Using the law of Identity, we get x+xy+w=x +w

Putting x=1,we get x + xy + w=1 + w.

Using the law of Dominance, we get x+xy+w=1

b) if w, x, and y be Boolean variables, where the value of x is 1, we need to determine the value of the expression xy+w and if it is not possible to find the value of the expression, then find the number of assignments of value for w and y that will result in the value 1 for the expression.
Putting x =1 in the given expression, we get xy+w=1•y+w. Using the law of Identity, we get xy+w=1+ w
Using the law of Dominance, we get xy+w=1 c) if w, x, and y be Boolean variables, where the value of x is 1, we need to determine the value of the expression xy+xw and if it is not possible to find the value of the expression, then find the number of assignments of value for w and y that will result in the value 1 for the expression.

Since x = 1 , we have x=0. Putting the value of x=1 and x=0.we get xy + xw = 0 *y + 1*w.

Using the law of Dominance in first term, we get xy + xw =0+1*w

Using the law of Identity, we get xy+xw=0+w *

Again, using the law of Identity, we get
0+w=w-

The number of assignment of value for w that will result the value 1 for the expression is only 1.
Therefore, we have

W=1

d) if w, x, and y be Boolean variables, where the value of x is 1, we need to determine the value of the expression xy+w and if it is not possible to find the value of the expression, then find the number of assignments of value for w and y that will result in the value 1 for the expression.
Since x = 1 ,we have x=0.

Putting the value of x =0, we get xy+w=0*y+w.

Using the law of Dominance, we get xy + w =0 + w -

Using the law of Identity, we get xy+w=w The number of assignment of value for w that will result the value 1 for the expression is only 1.

Therefore, we have w=1 15.2 Question 5

5. For the network in Fig. 15.7, express f as a function of w,x,y,z

w, x, y, z. f=w x yz+w x yz
=w x yz+z
=w x y

15.2 Question 8
8. For each of the following Boolean functions f , design a two-level gating network for f as a minimal sum of products.
a) f : B3 →B, where f (x, y, z) = 1 if and only if exactly two of the variables have the value 1.
b) f : B4 →B, where f (w, x, y, z) = 1 if and only if an odd number of variables have the value 1.

a) For Boolean function f which is defined as f→ B3→ B, where f(x, y. z) =1 if and only if exactly two of the variables have the value I ,we need to design a two level getting network for f as minimal sum of products.

If two variable have value 1 then possible cases are xyz , xyz, and xyz.

Hence .a two level getting network foif as minimal sum of products is xyz , xyz, and xyz.

b)For Boolean function f which is defined as f : B4 -> B, where f(w,x,y,z) = 1 if and only if odd number of variables have the value 1, we need to design a two level getting network for f as minimal sum of products

If two vamiable have value 1 then possible cases are wx yz, w x y z, w x y z, wx y z,w x y z, wx yz, wxyz, wxyz

Hence ,a two level getting network forf as minimal sum of products is wx yz + w x y z+ w x y z + wx y z+w x y z+ wx yz+ wxyz + wxyz