Download Document

Boolean Logic
ITI 1121
N. El Kadri
What is a switching network?
X1
X2
Xm
Switching
Network
Z1
Z2
Zm
Combinatorial Network: A stateless network. The
output is completely determined by the values of the input.
Sequential Network: The network stores an internal
state. The output is determined by the input,
and by the internal state.
2
Logic Functions: Boolean Algebra
INVERTER
X
X’
C=A·B
B
0
1
0
1
C
0
0
0
1
If A=1 AND B=1 then C=1
otherwise C=0
C=A+B
A
0
0
1
1
B
0
1
0
1
C
0
1
1
1
If A=1 OR B=1 then C=1
otherwise C=0
OR
A
B
If X=0 then X’=1
If X=1 then X’=0
A
0
0
1
1
AND
A
B
X’
1
0
X
0
1
3
Boolean expressions and logic
circuits
Any Boolean expression can be implemented as a logic circuit.
X = [A(C+D)]’+BE
C
D
C+D
A(C+D)
[A(C+D)]’
[A(C+D)]’+BE
A
B
E
BE
4
Basic Theorems: Operations with
0 and 1
X+0 = X
X
0
X+1 = 1
C=X
X
0
1
0
0
0
C
0
1
X
1
1
1
1
C
1
1
C=X
X
0
1
1
1
1
C
0
1
X·1 = X
X·0 = 0
X
0
C=1
X
0
1
C=0
X
0
1
0
0
0
C
0
0
X
1
5
Basic Theorems:
Idempotent Laws
X+X = X
X
X
C=X
X
0
1
X
0
1
C
0
1
C=X
X
0
1
X
0
1
C
0
1
X·X = X
X
X
6
Basic Theorems:
Involution Law
(X’)’=X
X
B
C=X
X
0
1
B
1
0
C
0
1
7
Basic Theorems:
Laws of Complementarity
X+X’ = 1
X
X’
C=1
X
0
1
X’
1
0
C
1
1
C=0
X
0
1
X’
1
0
C
0
0
X·X’ = 0
X
X’
8
Expression Simplification using
the Basic Theorems
X can be an arbitrarily complex expression.
Simplify the following boolean expressions as
much as you can using the basic theorems.
(AB’ + D)E + 1 = 1
(AB’ + D)(AB’ + D)’ = 0
(AB + CD) + (CD + A) + (AB + CD)’ = 1
9
Associative Law
(X+Y)+Z = X+(Y+Z)
X
Y
C
Z
X
0
0
0
0
1
1
1
1
Y
0
0
1
1
0
0
1
1
Z
0
1
0
1
0
1
0
1
X+Y
0
0
1
1
1
1
1
1
Y
Z
(X+Y)+Z
0
1
1
1
1
1
1
1
X
Y+Z
0
1
1
1
0
1
1
1
C
X+(Y+Z)
0
1
1
1
1
1
1
1
10
Associative Law
(XY)Z = X(YZ)
X
Y
C
Z
X
0
0
0
0
1
1
1
1
Y
Z
Y
0
0
1
1
0
0
1
1
Z
0
1
0
1
0
1
0
1
XY
0
0
0
1
0
0
1
1
(XY)Z
0
0
0
0
0
0
0
1
X
YZ
0
0
0
1
0
0
0
1
C
X(YZ)
0
0
0
0
0
0
0
1
11
First Distributive Law
X(Y+Z) = XY+XZ
X
0
0
0
0
1
1
1
1
Y
0
0
1
1
0
0
1
1
Z
0
1
0
1
0
1
0
1
Y+Z
0
1
1
1
0
1
1
1
X(Y+Z)
0
0
0
0
0
1
1
1
XY
0
0
0
0
0
0
1
1
XZ
0
0
0
0
0
1
0
1
XY+XZ
0
0
0
0
0
1
1
1
12
First Distributive Law
X(Y+Z) = XY+XZ
X
0
0
0
0
1
1
1
1
Y
0
0
1
1
0
0
1
1
Z
0
1
0
1
0
1
0
1
Y+Z
0
1
1
1
0
1
1
1
X(Y+Z)
0
0
0
0
0
1
1
1
XY
0
0
0
0
0
0
1
1
XZ
0
0
0
0
0
1
0
1
XY+XZ
0
0
0
0
0
1
1
1
13
First Distributive Law
X(Y+Z) = XY+XZ
X
0
0
0
0
1
1
1
1
Y
0
0
1
1
0
0
1
1
Z
0
1
0
1
0
1
0
1
Y+Z
0
1
1
1
0
1
1
1
X(Y+Z)
0
0
0
0
0
1
1
1
XY
0
0
0
0
0
0
1
1
XZ
0
0
0
0
0
1
0
1
XY+XZ
0
0
0
0
0
1
1
1
14
First Distributive Law
X(Y+Z) = XY+XZ
X
0
0
0
0
1
1
1
1
Y
0
0
1
1
0
0
1
1
Z
0
1
0
1
0
1
0
1
Y+Z
0
1
1
1
0
1
1
1
X(Y+Z)
0
0
0
0
0
1
1
1
XY
0
0
0
0
0
0
1
1
XZ
0
0
0
0
0
1
0
1
XY+XZ
0
0
0
0
0
1
1
1
15
First Distributive Law
X(Y+Z) = XY+XZ
X
0
0
0
0
1
1
1
1
Y
0
0
1
1
0
0
1
1
Z
0
1
0
1
0
1
0
1
Y+Z
0
1
1
1
0
1
1
1
X(Y+Z)
0
0
0
0
0
1
1
1
XY
0
0
0
0
0
0
1
1
XZ
0
0
0
0
0
1
0
1
XY+XZ
0
0
0
0
0
1
1
1
16
Second Distributive Law
X+YZ = (X+Y)(X+Z)
X
0
0
0
0
1
1
1
1
Y
0
0
1
1
0
0
1
1
Z
0
1
0
1
0
1
0
1
YZ
0
0
0
1
0
0
0
1
X+YZ
0
0
0
1
1
1
1
1
X+Y
0
0
1
1
1
1
1
1
X+Z
0
1
0
1
1
1
1
1
(X+Y)(X+Z)
0
0
0
1
1
1
1
1
17
Second Distributive Law
X+YZ = (X+Y)(X+Z)
X
0
0
0
0
1
1
1
1
Y
0
0
1
1
0
0
1
1
Z
0
1
0
1
0
1
0
1
YZ
0
0
0
1
0
0
0
1
X+YZ
0
0
0
1
1
1
1
1
X+Y
0
0
1
1
1
1
1
1
X+Z
0
1
0
1
1
1
1
1
(X+Y)(X+Z)
0
0
0
1
1
1
1
1
18
Second Distributive Law
(A different proof)
(X + Y)(X + Z) = X(X + Z) + Y(X + Z)
(using the first distributive law)
= XX + XZ + YX + YZ
(using the first distributive law)
= X + XZ + YX + YZ
(using the idempotent law)
= X·1 + XZ + YX + YZ
(using the operation with 1 law)
= X(1 + Z + Y) + YZ
(using the first distributive law)
= X·1 + YZ
(using the operation with 1 law)
= X + YZ
(using the operation with 1 law)
19
Simplification Theorems
XY + XY’ = X
XY + XY’ = X(Y + Y’) = X·1 = X
X + XY = X
X(1 + Y) = X·1 = X
(X + Y’)Y = XY
XY + Y’Y = XY + 0 = XY
(X + Y)(X + Y’) = X
(X + Y)(X + Y’) = XX + XY’ + YX + YY’
= X + X(Y’ + Y) + 0
= X + X·1
=X
X(X + Y) = X
X(X + Y) = XX + XY = X·1 + XY
= X(1 + Y) = X·1 = X
XY’ + Y = X + Y
(using the second distributive law)
XY’ + Y = Y + XY’ = (Y + X)(Y + Y’)
= (Y + X)·1 = X + Y
20
Examples
Simplify the following expressions:
W = [M + N’P + (R + ST)’][M + N’P + R + ST]
X = M + N’P
Y = R + ST
W = (X + Y’)(X + Y)
W = XX + XY + Y’X + Y’Y
W = X·1 + XY + XY’ + 0
W = X + X(Y + Y’) = X + X·1 = X
W = M + N’P
21