DeMorgan’s Theorem uses two sets of rules or laws to solve various Boolean algebra expressions by changing OR’s to AND’s, and AND’s to OR’s
Boolean Algebra uses a set of laws and rules to define the operation of a digital logic circuit with “0’s” and “1’s” being used to represent a digital input or output condition. Boolean Algebra uses these zeros and ones to create truth tables and mathematical expressions to define the digital operation of a logic AND, OR and NOT (or inversion) operations as well as ways of expressing other logical operations such as the XOR (Exclusive-OR) function.
While George Boole’s set of laws and rules allows us to analyise and simplify a digital circuit, there are two laws within his set that are attributed to Augustus DeMorgan (a nineteenth century English mathematician) which views the logical NAND and NOR operations as separate NOT AND and NOT OR functions respectively.
But before we look at DeMorgan’s Theory in more detail, let’s remind ourselves of the basic logical operations where A and B are logic (or Boolean) input binary variables, and whose values can only be either “0” or “1” producing four possible input combinations, 00, 01, 10, and 11.
Truth Table for Each Logical Operation
Input Variable | Output Conditions |
A | B | AND | NAND | OR | NOR |
0 | 0 | 0 | 1 | 0 | 1 |
0 | 1 | 0 | 1 | 1 | 0 |
1 | 0 | 0 | 1 | 1 | 0 |
1 | 1 | 1 | 0 | 1 | 0 |
The following table gives a list of the common logic functions and their equivalent Boolean notation where a “.” (a dot) means an AND (product) operation, a “+” (plus sign) means an OR (sum) operation, and the complement or inverse of a variable is indicated by a bar over the variable.
Logic Function | Boolean Notation |
AND | A.B |
OR | A+B |
NOT | A |
NAND | A .B |
NOR | A+B |
DeMorgan’s Theory
DeMorgan’s Theorems are basically two sets of rules or laws developed from the Boolean expressions for AND, OR and NOT using two input variables, A and B. These two rules or theorems allow the input variables to be negated and converted from one form of a Boolean function into an opposite form.
DeMorgan’s first theorem states that two (or more) variables NOR´ed together is the same as the two variables inverted (Complement) and AND´ed, while the second theorem states that two (or more) variables NAND´ed together is the same as the two terms inverted (Complement) and OR´ed. That is replace all the OR operators with AND operators, or all the AND operators with an OR operators.
DeMorgan’s First Theorem
DeMorgan’s First theorem proves that when two (or more) input variables are AND’ed and negated, they are equivalent to the OR of the complements of the individual variables. Thus the equivalent of the NAND function will be a negative-OR function, proving that A.B = A+B. We can show this operation using the following table.
Verifying DeMorgan’s First Theorem using Truth Table
Inputs | Truth Table Outputs For Each Term |
B | A | A.B | A.B | A | B | A + B |
0 | 0 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 0 | 0 | 0 | 0 |
We can also show that A.B = A+B using logic gates as shown.
DeMorgan’s First Law Implementation using Logic Gates
The top logic gate arrangement of: A.B can be implemented using a standard NAND gate with inputs A and B. The lower logic gate arrangement first inverts the two inputs producing A and B. These then become the inputs to the OR gate. Therefore the output from the OR gate becomes: A+B
Then we can see here that a standard OR gate function with inverters (NOT gates) on each of its inputs is equivalent to a NAND gate function. So an individual NAND gate can be represented in this way as the equivalency of a NAND gate is a negative-OR.
DeMorgan’s Second Theorem
DeMorgan’s Second theorem proves that when two (or more) input variables are OR’ed and negated, they are equivalent to the AND of the complements of the individual variables. Thus the equivalent of the NOR function is a negative-AND function proving that A+B = A.B, and again we can show operation this using the following truth table.
Verifying DeMorgan’s Second Theorem using Truth Table
Inputs | Truth Table Outputs For Each Term |
B | A | A+B | A+B | A | B | A . B |
0 | 0 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 0 | 1 | 0 |
1 | 0 | 1 | 0 | 1 | 0 | 0 |
1 | 1 | 1 | 0 | 0 | 0 | 0 |
We can also show that A+B = A.B using the following logic gates example.
DeMorgan’s Second Law Implementation using Logic Gates
The top logic gate arrangement of: A+B can be implemented using a standard NOR gate function using inputs A and B. The lower logic gate arrangement first inverts the two inputs, thus producing A and B. Thus then become the inputs to the AND gate. Therefore the output from the AND gate becomes: A.B
Then we can see that a standard AND gate function with inverters (NOT gates) on each of its inputs produces an equivalent output condition to a standard NOR gate function, and an individual NOR gate can be represented in this way as the equivalency of a NOR gate is a negative-AND.
Although we have used DeMorgan’s theorems with only two input variables A and B, they are equally valid for use with three, four or more input variable expressions, for example:
For a 3-variable input
A.B.C = A+B+C
and also
A+B+C = A.B.C
For a 4-variable input
A.B.C.D = A+B+C+D
and also
A+B+C+D = A.B.C.D
and so on.
DeMorgan’s Equivalent Gates
We have seen here that by using DeMorgan’s Theorems we can replace all of the AND (.) operators with an OR (+) and vice versa, and then complements each of the terms or variables in the expression by inverting it, that is 0’s to 1’s and 1’s to 0’s before inverting the entire function.
Thus to obtain the DeMorgan equivalent for an AND, NAND, OR or NOR gate, we simply add inverters (NOT-gates) to all inputs and outputs and change an AND symbol to an OR symbol or change an OR symbol to an AND symbol as shown in the following table.
DeMorgan’s Equivalent Gates
Standard Logic Gate | DeMorgan’s Equivalent Gate |
| |
| |
| |
| |
Then we have seen in this tutorial about DeMorgan’s Thereom that the complement of two (or more) AND’ed input variables is equivalent to the OR of the complements of these variables, and that the complement of two (or more) OR’ed variables is equivalent to the AND of the complements of the variables as defined by DeMorgan.
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Hey there,
Electronics Guru! I stumbled upon your article on De Morgan’s Theorem, and I have to say, it’s like you’ve cracked the code to demystify Boolean algebra. Your explanations were crystal clear, especially for someone like me who’s still wrapping their head around these concepts.
I’ve been struggling to understand how De Morgan’s Theorem fits into the bigger picture of digital logic, but your breakdown made it feel like a light bulb went off in my brain. The way you illustrated how to simplify complex Boolean expressions using both the AND and OR gates drove the point home. It’s amazing how such seemingly intricate rules can simplify logic circuits so effectively. Thanks to your article, I feel much more confident diving deeper into the world of digital electronics.
Looking forward to more insights from you!
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Well..explaination and understood but how to proved demorgan’s law by using algebraically?
Well Boolean algebra is not what I would call “Algebra.” If you’re asking for a proof, I would start by making two logic systems which are the same according to demorgans. Make tables… The best way to learn is by examples. I have never Truley done the math. I just memorized the theorems based off examples I made for myself.
You should include a tutorial on K Maps.
Write logical diagram of de morgan
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The first law of Demorgan theorem is (complement the given expression to remove the overall NOT sign) Then how are we going to do this kind question (j+kl). Since it does not carry not sigh
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