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DeMorgan's
Theorem
Rules of Boolean :
DeMorgan's 1st Theorem :
The complement of a product of variables is
equal to the sum of the complemented
variables.
AB = A + B
Applying DeMorgan's 1st Theorem to gates:
A B AB A+B
0 0 1 1
0 1 1 1
1 0 1 1
1 1 0 0
DeMorgan's 2st Theorem :
The complement of a sum of variables is equal to
the product of the complemented variables.
A + B = A B
Applying DeMorgan's 2st Theorem to gates:
A B AB A+B
0 0 1 1
0 1 0 0
1 0 0 0
1 1 0 0
A B
Example:
Apply DeMorgan's theorem to remove the over bar
covering both terms from the expression X= C + D.
X= C . D
X= C . D
X= C . D
Individual Work
SOP and POS forms:
Boolean expressions can be written in the sum-of-product form (SOP) or
in the product-of-sum (POS). These forms can simplify the
implementation of combinational logic. In both forms, an over-bar cannot
extend over more than one variable.
An expression is in SOP form when two or more product terms are summed as in
the following examples:
A B C + A B A B C + C D C D + E
An expression is in POS form when two or more product terms are multiplied as
in the following examples:
(A + B) (A + C) (A + B + C) (B + D) (A + B) C
 Sum of Products (SOP)
ABCCBACBACBAF 
AC
BBAC

 )(
CB
AACB

 )(
BA
BA
CCBA



)1(
)(
)()()( BBACCCBAAACBF 
ACBACBF 
 Product of Sums (POS)
)( AACB 
)( BBCA 
)( CCBA 
)()()( AACBCCBABBCAF 
CBBACAF 
CABBCACBACBAF 
))()(( CBBACAF 
Group work
Question
One of De Morgan's theorems states that . Simply
stated, this means that logically there is no
difference between:
A. a NOR and an AND gate with inverted inputs
B. a NAND and an OR gate with inverted inputs
C. an AND and a NOR gate with inverted inputs
D .a NOR and a NAND gate with inverted inputs
Question
Which of the following expressions is in the sum-
of-products (SOP) form?
A. (A + B)(C + D)
B. (A)B(CD)
C. AB(CD)
D. AB + CD
Question
Use Boolean algebra to find the most simplified SOP
expression for F = ABD + CD + ACD +ABC + ABCD.
A. F = ABD + ABC + CD
B. F = CD + AD
C. F = BC + AB
D. F = AC + AD
Question
The commutative law of Boolean addition states
that A + B = A × B.
A. True
B. False

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DeMorgan's Theorem Rules and Applications

  • 1.
  • 4. DeMorgan's 1st Theorem : The complement of a product of variables is equal to the sum of the complemented variables. AB = A + B Applying DeMorgan's 1st Theorem to gates: A B AB A+B 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0 0
  • 5. DeMorgan's 2st Theorem : The complement of a sum of variables is equal to the product of the complemented variables. A + B = A B Applying DeMorgan's 2st Theorem to gates: A B AB A+B 0 0 1 1 0 1 0 0 1 0 0 0 1 1 0 0 A B
  • 6. Example: Apply DeMorgan's theorem to remove the over bar covering both terms from the expression X= C + D. X= C . D X= C . D X= C . D
  • 8. SOP and POS forms: Boolean expressions can be written in the sum-of-product form (SOP) or in the product-of-sum (POS). These forms can simplify the implementation of combinational logic. In both forms, an over-bar cannot extend over more than one variable. An expression is in SOP form when two or more product terms are summed as in the following examples: A B C + A B A B C + C D C D + E An expression is in POS form when two or more product terms are multiplied as in the following examples: (A + B) (A + C) (A + B + C) (B + D) (A + B) C
  • 9.  Sum of Products (SOP) ABCCBACBACBAF  AC BBAC   )( CB AACB   )( BA BA CCBA    )1( )( )()()( BBACCCBAAACBF  ACBACBF 
  • 10.  Product of Sums (POS) )( AACB  )( BBCA  )( CCBA  )()()( AACBCCBABBCAF  CBBACAF  CABBCACBACBAF  ))()(( CBBACAF 
  • 12.
  • 13. Question One of De Morgan's theorems states that . Simply stated, this means that logically there is no difference between: A. a NOR and an AND gate with inverted inputs B. a NAND and an OR gate with inverted inputs C. an AND and a NOR gate with inverted inputs D .a NOR and a NAND gate with inverted inputs
  • 14. Question Which of the following expressions is in the sum- of-products (SOP) form? A. (A + B)(C + D) B. (A)B(CD) C. AB(CD) D. AB + CD
  • 15. Question Use Boolean algebra to find the most simplified SOP expression for F = ABD + CD + ACD +ABC + ABCD. A. F = ABD + ABC + CD B. F = CD + AD C. F = BC + AB D. F = AC + AD
  • 16. Question The commutative law of Boolean addition states that A + B = A × B. A. True B. False