2.1.5.
AK DeMorgan’s Theorems
Introduction
Despite all of the work done by George Boole, there was still more work to be done.
Expanding on Boole’s studies, Augustus DeMorgan (1806-1871) developed two additional
theorems that now bear his name. Without DeMorgan’s Theorems, the complete
simplification of logic expression would not be possible.
Theorem #1: X Y X Y
Theorem #2 : X Y X Y
As we will see in later activities, DeMorgan’s Theorems are the foundation for the NAND and
NOR logic gates.
In this activity you will learn how to simplify logic expressions and digital logic circuits using
DeMorgan’s two theorems along with the other laws of Boolean algebra.
Procedure
Using DeMorgan’s theorems and the other theorems and laws of Boolean algebra, simplify
the following logic expressions. Note the theorem/law used at each simplification step. Be
sure to put your answer in Sum-Of-Products (SOP) form.
Rules Applied
1.
F 1= X +Y 14 B (DM ) ; 9 (Boole)
F 1= X Y
2.
F 2 =( R+S )(T +U ) 14 A (DM)
F 2 =( R+S )+(T +U ) 14 B, 14 B (DM)
F 2 =( R S )+(T U ) 9 (Boole)
F 2 =( R S )+(T U )
3.
F 3=M N 14 A (DM)
F 3=M +N 9 (Boole)
F 3=M +N
© 2014 Project Lead The Way, Inc.
Digital Electronics ANSWER KEY 2.1.5 DeMorgan’s Theorems – Page 1
4.
F 4 =W X (Y +Z ) 14 A (DM)
F 4 =W X +(Y +Z ) 14 A, 14 B (DM)
F 4 =W +X +(Y Z ) 14 A (DM), 9 (Boole)
F 4 =W +X +Y Z
5.
F 5 =P Q R+ PQ R 14 B (DM)
F 5 =( P Q R)( P Q R) 9 (Boole)
F 5 =( P Q R)( P Q R ) 12 A (DM)
F 5 =P PQ QR R 4,4,4 (Boole)
F 5 =( 0)+(0)+(0) 5 (Boole)
F 5 =0
6.
F 6 =W X Y Z 14 A (DM)
9, 9 (Boole)
F 6 =W X Y +Z
F 6 =W X Y +Z 14 A (DM)
F 6 =(W +X )Y +Z 9 (Boole)
F 6 =(W +X )Y +Z 12A (DM)
F 6 =WY + X Y +Z
Let’s see how we would utilize DeMorgan’s theorems to simplify a digital logic circuit.
7. Write the UN-SIMPLIFIED logic expression for the output Do-Nothing in the logic
circuit shown below.
© 2014 Project Lead The Way, Inc.
Digital Electronics ANSWER KEY 2.1.5 DeMorgan’s Theorems – Page 2
The output is call Do-Nothing because that is exactly what the circuit does, nothing; yet it’s a
good example for learning about DeMorgan’s theorems.
8. Using DeMorgan’s theorems and the other theorems and laws of Boolean algebra,
simplify the logic expression Do-Nothing. Be sure to put your answer in Sum-Of-
Products (SOP) form.
DN=(Y Z )( X +Y )+( X+Y ) 14 A, 14 B, 14 B (DM)
DN=(Y +Z )( X Y )+( X Y ) 9,9,9, (Boole)
DN=(Y +Z )( X Y )+( X Y ) 12 A (DM)
DN=X Y Y +X Y Z +X Y 4 (Boole)
DN=0+ X Y Z+ X Y 5 (Boole)
DN=X Y Z + X Y
9. In the space provided, draw an AOI circuit that implements the simplified logic
expression Do-Nothing. For this implementation you may assume that AND & OR
gates are available with any number of inputs.
© 2014 Project Lead The Way, Inc.
Digital Electronics ANSWER KEY 2.1.5 DeMorgan’s Theorems – Page 3
Do Nothing – I
10. Re-implement the circuit assuming that only 2-input AND gates (74LS08), 2-input OR
gates (74LS32), and inverters (74LS04) are available. Draw this circuit in the space
provided.
Do Nothing – II
Conclusion
1. Draw the gate equivalent for DeMorgan’s two theorems.
Theorem #1 : X⋅Y = X +Y
X
X
X Y
X
X Y X Y
Y Y
Y
Theorem #2 : X +Y =X⋅Y
X
X
X Y
X X Y
X Y Y
Y
Y
2. How would you prove that the original Do-Nothing circuit and the simplified version are
equivalent?
Compare the truth tables.
3. If each GATE cost 5¢ and you made 100,000 of the unsimplified units, how much of the
company’s money did you waste on the unsimplified Do-Nothing project?
Do Nothing Un-simplified Do Nothing Simplified
A = 2 Gates (2-input) A = 3 Gates (2-input)
O = 3 Gates (2-input) O = 1 Gates (2-input)
I = 6 Gates (2-input) I = 2 Gates (2-input)
11 GATES x (.05) x 100,000 = $55,000 6 GATES x (.05) x 100,000 = $30,000
© 2014 Project Lead The Way, Inc.
Digital Electronics ANSWER KEY 2.1.5 DeMorgan’s Theorems – Page 4
You would have wasted $25,000.
© 2014 Project Lead The Way, Inc.
Digital Electronics ANSWER KEY 2.1.5 DeMorgan’s Theorems – Page 5
