If you happen to be viewing the article What is De Morgan’s First Law?? on the website Math Hello Kitty, there are a couple of convenient ways for you to navigate through the content. You have the option to simply scroll down and leisurely read each section at your own pace. Alternatively, if you’re in a rush or looking for specific information, you can swiftly click on the table of contents provided. This will instantly direct you to the exact section that contains the information you need most urgently.
The students at Vedantu will learn about the concept of De Morgan’s First Law in detail. They are provided with an entire guide to the topic to help them learn the basic concepts in depth. The website of Vedantu provides the students with lots of interesting and engaging study material including the concepts from Classes 1 to 12, revision notes, worksheets, sample papers, previous year’s question papers, NCERT Exemplar, and NCERT textbooks along with their solutions to clear the doubts of the students. Moreover, the students can refer to them easily without any hassle for free of cost at the time of their convenience.
The concepts discussed here have been prepared by the subject experts at Vedantu who have years of experience and expertise in the respective fields. Also, the topics have been covered as per the latest syllabus of the concerned boards to help the students get all the resources in one place without any difficulty. The resources and study material provided at Vedantu are 100% accurate and reliable for the use by the students to score good grades in their exams and life.
In algebra, De Morgan’s First Law or First Condition states that the complement of the product of two variables is corresponding to the sum of the complement of each variable. In other words, according to De-Morgan’s first Laws or first theorem if ‘A’ and ‘B’ are the two variables or Boolean numbers. This indicates that the NAND gate function is similar to the OR gate function with complemented inputs. Then accordingly the equations are as below;-
For NOR Gate
Y=A+B=Y=A¯+B¯
For the Bubbled AND Gate
Y=A.B=Y=A¯.B¯
Contents
Symbolic representation of De Morgan’s First Law Theorem
Since the NOR and the bubbled gates can be interchanged, i.e., both gates have just similar outputs for the identical set of inputs.
Hence, the equation can be algebraically represented in the figure shown below.
This equation presented above is known as DeMorgan’s First Theorem. The symbolic illustration of the theorem is presented as shown below.
Role of Complementation Bars
Complementation bars are proposed to operate as grouping symbols. Hence, when a bar is broken, the expressions beneath it should remain grouped. Parentheses may be positioned around these grouped expressions as assistance to give a miss to changing precedence.
Verifying DeMorgan’s First Theorem Using Truth Table
According to DeMorgan’s First Law, it proves that in conditions where two (or more) input variables are Added and negated, they are equal to the OR of the complements of the separate variables. Hence, the equivalent of the NAND function and is a negative-OR function verifying that A.B = A+B and we can prove this using the following table.
DeMorgan’s First Theorem Proof using Truth Table
Now that you have already understood DeMorgan’s First Theorem using the Truth Table. We will make you familiar with another way to prove the theorem i.e. by using logic gates.
This is to say, we can also prove that A.B = A+B using logic gates as hereinafter.
Verifying and Execution of DeMorgan’s First Law using Logic Gates
The uppermost logic gate placement of: A.B can be executed considering a NAND gate with inputs A and B. The lowermost logic gate placement, in the beginning, inverts the two inputs yielding A and B which become the inputs to the OR gate. Thus, with this, the output from the OR gate becomes A+B.
Therefore, an OR gate with inverters (NOT gates) on its every input is equal to a NAND gate function, and an independent NAND gate can be showcased in this way the equality of a NAND gate is a negative – OR.
Simplifying DeMorgan’s First Law with Example
According to DeMorgan’s First Law, what is an equivalent statement to “The kitchen floor needs mopping and the utensils need washing, but I will not do both.”?
The two postulations are “I will mop the kitchen floor” and “I will wash the utensils.” Simply modify the given statement to an “OR” statement and negate each of these postulations:
“Either I will not mop the kitchen floor or I will not wash the utensils.”
P.S: that this statement lays open the likelihood that one of the tasks is completed, and it is also possible that neither chore is being completed.
Importance of De Morgan’s Law:
De Morgan’s first Law holds an important place in the syllabus of the students of higher secondary Classes, as it checks on a few crucial topics, it’s important features are highlighted below as:
-
The theorems of De Morgan’s Law have been proved to be very useful for simplifying Boolean logic expressions due to the way they can ‘break’ an inversion, which could be the complement of a complex Boolean expression.
-
The theorems of De Morgan’s can also be used to express logic expressions that do not originally contain inversion terms differently. Also, this can again prove to be useful while simplifying the Boolean equations. When used in this way it must be taken care of to not to ‘forget’ the final inversion, which can be easily avoided by complementing both the sides of the expression to be simplified before applying De Morgan’s theorem and then again complementing after simplification.
-
Lastly, the students must note that one way of interpreting De Morgan’s theorem is that any AND/OR operation can be considered as an OR/AND operation as long as NOT gates are used as well in the equation for ease of calculation.
Solved Examples
Problem1:
How to deduce the following equation to standard form?
F = MNO +M’N
F’ = (MNO + M’N)’l
Solution1:
Using the De Morgan’s Law
We get,
= (MNO)’ (M’N)’
= (M’+N’+O’) (M+N’)
Now, applying the Law of distributivity
= N’ + (M’+O’) M
Again, applying Distributivity
= N’ + M’M + OEM
= N’ + MO’ (standard form)l
Problem2:
Apply De Morgan’s Law to determine the inverse of the below given equation and reduce to the form of the sum-of-product:
F = MO’ + MNP’ + MOP
Solution2:
F’= (MO’ + MNP’ + MOP)’
= (MO’)’ (MNP’)’ (MOP)’
= (M’+O)(M’+N’+P)(M’+O’+P’)
= M’+O (N’+P) (O’+P’)
= M’+ (N’+P) OP’
= M’ + ON’P’ + OPP’
Thus, we get = M’ + ON’P.
Fun Facts
-
Do you know the full form of DeMorgan’s Theorems? It’s Demorgan’s theorem.
-
No matter whether De Morgan’s Laws apply to sets, propositions, or logic gates, the anatomy always remains the same.
Thank you so much for taking the time to read the article titled What is De Morgan’s First Law? written by Math Hello Kitty. Your support means a lot to us! We are glad that you found this article useful. If you have any feedback or thoughts, we would love to hear from you. Don’t forget to leave a comment and review on our website to help introduce it to others. Once again, we sincerely appreciate your support and thank you for being a valued reader!
Source: Math Hello Kitty
Categories: Math
