Basic Identities of Boolean Algebra
Boolean Algebra
Chapter 2-3
Z. Aliyazici
14.
15.
16.
17.
Ex:
X(Y+Z) = XY+XZ (Distributed)
X+(YZ) = (X+Y)(X+Z)
(X+Y)’ = X’.Y’
(DeMorgan’s)
(X.Y)’ = X’+Y’
AB +C+1 = 1
X+(Y+Z) = (X+Y)+Z = X+Y+Z
X(YZ) = (XY)Z = XYZ
Examples:
•
•
•
•
•
•
XY+XY’=X(Y+Y’)=X
(X+Y)(X+Y’)=X+(YY’)=X
X+XY’=X(1+Y’)=X
X(X+Y)=X+XY=X(1+Y)=X
(X+Y’)Y=XY+Y’Y=XY
XY’+Y=(X+Y)(Y+Y’)=(X+Y)1=X+Y
1. X+0 = X
2. X.1 = X
3. X+1 = 1
4. X.0 = 0
5. X+X = X
6. X.X = X
7. X+X’ = 1
8. X.X’ = 0
9. (X’)’=X
10. X+Y = Y+X
11. XY = YX (Commutative)
12. X+(Y+Z) = (X+Y)+Z
(Associative)
13. X(YZ) = (XY)Z
Ex: Show that
a. (A+B)(A+CD) = A + BCD
= A+ACD+BA+BCD
= A(1+CD+B)+BCD
= A1+BCD
= A+BCD
b. (X+Y)(X+Z) = XX+XZ+XY+YZ
= X+XZ+XY+YZ
= X(1+Z+Y)+YZ
= X+YZ
Examples:
Z=[A+B’C+D+EF][A+B’C+(D+EF)’]
Z=[ X + Y ][ X + Y’
]
Z=X=A+B’C
• Z=A’BC+A’=A’(BC+1)=A’.1=A’
1
Standard Form
Sum of product
• A Boolean function can be expressed in
a different algebraic ways.
• The standard forms contain product
terms and sum term
Example: F=XYZ is in product form use AND
operations
F=(X+Y)(X+Z)Z is in product form
F=X+Y+Z is in logical sum form use OR
gate
F=XY+XZ+ZY is in sum form
Sum of product form: all product are the
product of single variables only
Example:
AB’+CD’E+AC’E’
ABC’+DEFG+H
A
B
C
D
E
A
C
E
Product of Sums
•
•
•
•
•
•
Example
All sums are the sums of single variable
Example:
(A+B’)(C+D’+E)(A+C’+E’)
(A+B)(C+D+E)F
AB’C(D’+E)
Considered product of sums
(A+BC)(A+D+E)
=A+AD+AE+ABC+BCD+BCE
=A(1+D+E+BC)+BCD+BCE
Product of sum
=A+BCD+BCE
Sum of product
Example:
AB’+C’D
= (AB’+C’)(AB’+D)
= (A+C’)(B’+C’)(A+D)(B’+D)
Sum of product
Product of sum
Algebraic Manipulation
X’
X
F = X’YZ+X’YZ’+XZ
= X’Y(Z+Z’) +XZ
(by identity 14)
= X’Y.1+XZ
= X’Y + XZ
(by identity 7)
(by identity 2)
X’YZ
X’YZ’
Y
Z
F
Z’
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
F
0
0
1
1
0
1
0
1
F = X’YZ+X’YZ’+XZ
X
X’
X’ Y
Y
F=X’Y+XZ
Z
XZ
F=X’Y + XZ
2
E.g., prove the theorem:
X • Y + X • Y' = X
• X(X+Y)=X+XY=X
X • Y + X •Y‘ = X • (Y + Y')
= X • (1)
=X
E.g., prove the theorem:
X + X•Y
X + X•Y
• (X+Y)(X+Y’)=X+YY’+XY’+XY
=X+0+X(Y’+Y)=X+X(1)=X
= X?
=
=
=
=
• XY+X’Z+YZ=XY+X’Z ?
XY+X’Z+YZ
= XY+X’Z+YZ(X+X’)
= XY+X’Z+XYZ+X’YZ
= XY+XYZ+X’Z+X’YZ
= XY(1+Z)+X’Z(1+Y)
= XY+X’Z
X•1 + X•Y
X • (1 + Y)
X • (1)
X
Inversion
• (A+B)(A’+C) = AA’+AC+BA’+BC
= AC+A’B+BC
= AC+A’B
Using DEMorgan’s Laws
(X+Y)’ = X’.Y’
X
Y
0
0
1
1
0
1
0
1
X+Y (X+Y)’
0
1
1
1
1
0
0
0
X
Y
X’
Y’
X’.Y’
0
0
1
1
0
1
0
1
1
1
0
0
1
0
1
0
1
0
0
0
Inversion
Using DEMorgan’s Laws
(XY)’ = X’+Y’
X
Y
XY
(XY)’
X
Y
X’
Y’
X’+Y’
0
0
1
1
0
1
0
1
0
0
0
1
1
1
1
0
0
0
1
1
0
1
0
1
1
1
0
0
1
0
1
0
1
1
1
0
•
•
•
•
In general case, DeMorgen’s laws:
(X1X2…XN )’=X1’+X2 ’+…+Xn’
And
(X1+X2+…+Xn )’=X1’X2’…Xn ’
Example:Find the inverse (complement) of
(A’+B)C’
[(A’+B)C’]’=(A’+B)’+(C’)’
=(A’)’(B’)+C
=AB’+C
3
Complement Function
• Find the complement of the following function
F1 = [X’YZ’+X’Y’Z]’
= (X’YZ’)’.(X’Y’Z)’
= (X+Y’+Z)(X+Y+Z’)
F2 = [X(Y’Z’+YZ)]’
= X’+(Y’Z’+YZ)’
=X’+(Y’Z’)’(YZ)’
= X’+(Y+Z)(Y’+Z’)
=X’+YY’+YZ’+Y’Z+ZZ’
=X’+YZ’+Y’Z
Duality
• Given a Boolean expression, replace
AND with OR, OR with AND, 0 with 1,
ad 1 with 0.
• Example:
– F=AB’+C+0D’(1+E)
– FD=(A+B’)C(1+D’+0E)
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