Evaluate the logical expressions with x = y = 1, and w = z = 0. Evaluate the following expression xy + (x+w+yz)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Q3
D
Question 3
Using the given laws:
Table 7.1.4: Laws of Boolean algebra.
Idempotent laws:
Associative laws:
Commutative laws:
Complement laws:
De Morgan's laws:
Absorption laws:
X+X=X
Distributive laws:
Identity laws:
Domination laws:
Double complement law: =
xy + (x +w+yz)
(x + y) + z = x + (y+z) (xy)z = x(yz)
x+y=y+X
x+yz=(x+y)(x + 2)
X+0=X
x 0 = 0
XX=0
1=0
X*X=X
x+y=xy
x + (xy) = x
xy = yx
x(y + 2) = xy + xz
X. 1 = x
x+1=1
X+X=1
0 = 1
xy=x+y
x(x + y) = x
Evaluate the logical expressions with x = y = 1, and w = z = 0.
Evaluate the following expression
Transcribed Image Text:D Question 3 Using the given laws: Table 7.1.4: Laws of Boolean algebra. Idempotent laws: Associative laws: Commutative laws: Complement laws: De Morgan's laws: Absorption laws: X+X=X Distributive laws: Identity laws: Domination laws: Double complement law: = xy + (x +w+yz) (x + y) + z = x + (y+z) (xy)z = x(yz) x+y=y+X x+yz=(x+y)(x + 2) X+0=X x 0 = 0 XX=0 1=0 X*X=X x+y=xy x + (xy) = x xy = yx x(y + 2) = xy + xz X. 1 = x x+1=1 X+X=1 0 = 1 xy=x+y x(x + y) = x Evaluate the logical expressions with x = y = 1, and w = z = 0. Evaluate the following expression
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