What Is Demorgans Law?
De Morgan’s Law is a collection of boolean algebra transformation rules that are used to connect the intersection and union of sets using complements. De Morgan’s Law states that two conditions must be met. These conditions are typically used to simplify complex expressions.
This makes performing calculations and solving complicated boolean expressions easier. According to De Morgan’s Law logic, the complement of the union of two sets is equal to the intersection of their separate complements. Furthermore, the complement of two sets intersecting is equal to the sum of their separate complements.
Venn diagrams make it simple to visualise these laws. We will study Demorgan’s Law’s statements, how to prove them, how to apply them, and De Morgan’s law example in this article.
What are De Morgan’s Law explain?
De Morgans law : The complement of the union of two sets is the intersection of their complements and the complement of the intersection of two sets is the union of their complements. These are called De Morgans laws. These are named after the mathematician De Morgan.
What does De Morgan’s second law state?
What is De Morgan’s second law? Second law states that the complement of the intersection of two sets is the same as the union of their complements.
What is De Morgan’s theorem with example?
DeMorgan’s Theorem states that inverting the output of any gate results in same function as opposite type of gate (AND vs. OR) with two inverted variables A and B. It is used to solve Boolean Algebra expressions. It perfomes gate operation like NAND gate and NOR gate.
What is De Morgan’s first theorem?
De Morgan’s First Law 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.
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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,
- For NOR Gate
- Y=A+B=Y=A¯+B¯
- For the Bubbled AND Gate
Y=A.B=Y=A¯.B¯
Who created Demorgan’s law?
| Augustus De Morgan | |
|---|---|
| Born | 27 June 1806 Madurai, Carnatic, Madras Presidency, (present-day India) |
| Died | 18 March 1871 (aged 64) London, England |
| Nationality | British |
| Alma mater | Trinity College, Cambridge |
| Known for | De Morgan’s laws De Morgan algebra De Morgan hierarchy Relation algebra Universal algebra |
| Scientific career | |
| Fields | Mathematician and logician |
| Institutions | University College London University College School |
| Academic advisors | John Philips Higman George Peacock William Whewell |
| Notable students | Edward Routh James Joseph Sylvester Frederick Guthrie William Stanley Jevons Ada Lovelace Francis Guthrie Stephen Joseph Perry |
| Influences | George Boole |
| Influenced | Thomas Corwin Mendenhall Isaac Todhunter |
| Notes | |
| He was the father of William De Morgan, |
Augustus De Morgan (27 June 1806 – 18 March 1871) was a British mathematician and logician, He formulated De Morgan’s laws and introduced the term mathematical induction, making its idea rigorous.
What is De Morgan’s law formula?
De Morgan’s Law of Union: The complement of the union of the two sets A and B will be equal to the intersection of A’ (complement of A) and B’ (complement of B). This is also known as De Morgan’s Law of Union. It can be represented as (A ∪ B)’ = A’ ∩ B’.
Why is De Morgan’s Law Important?
De Morgan’s Laws describe how mathematical statements and concepts are related through their opposites. In set theory, De Morgan’s Laws relate the intersection and union of sets through complements. In propositional logic, De Morgan’s Laws relate conjunctions and disjunctions of propositions through negation.
What is De Morgan’s theorem formula?
De-Morgan’s First Theorem – According to the first theorem, the complement result of the AND operation is equal to the OR operation of the complement of that variable. Thus, it is equivalent to the NAND function and is a negative-OR function proving that (A.B)’ = A’+B’ and we can show this using the following table.
| Inputs | Output For Each Term | |||||
|---|---|---|---|---|---|---|
| A | B | A.B | (A.B)’ | A’ | B’ | A’A+B’ |
| 1 | 1 | 1 | 1 | |||
| 1 | 1 | 1 | 1 | |||
| 1 | 1 | 1 | 1 | |||
| 1 | 1 | 1 |
What are the two De Morgan’s theorem?
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.
How do you remember De Morgan’s law?
Fundamental Hardware Elements of Computers: De Morgan’s Laws – Wikibooks, open books for an open world De Morgan’s laws are used to simplify Boolean equations so that you can build equations only involving one sort of gate, generally only using NAND or NOR gates. This can lead to cheaper hardware. There are two laws that you need to remember:
| P, Q ¯ = P ¯ + Q ¯ }= }+ }} | P + Q ¯ = P ¯, Q ¯ }= }. }} | |
| Rule 1 | Rule 2 |
|---|
An easy way to remember De Morgan’s Laws is through the rhyme: “break the line, change the sign” ! Let’s prove that I’m not lying to you by creating a truth table to prove that: P + Q ¯ = P ¯, Q ¯ }= }. }} Answer:
| P | Q | P + Q + } | P + Q ¯ }} | P ¯ }} | Q ¯ }} | P ¯, Q ¯ }. }} |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 |
Since the values in the 4th and last columns are the same for all rows (which cover all possible truth value assignments to the variables), we can conclude that the two expressions are logically equivalent. Now we prove P, Q ¯ = P ¯ + Q ¯ }= }+ }} by the same method: Answer:
| P | Q | P,Q, } | P, Q ¯ }} | P ¯ }} | Q ¯ }} | P ¯ + Q ¯ }+ }} |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 |
There is a rather nice concrete way of thinking about this, with a gate that’s padlocked with two padlocks, padlock 1 and padlock 2. We’ll use P to stand for padlock 1 is open, and Q to stand for padlock 2 is open. You can go through the gate if padlock 1 is open AND padlock 2 is open ( P,
- Q ) You can not go through the gate if padlock 1 is locked OR padlock 2 is locked ( P ¯ + Q ¯ }+ }} ) Since ‘You can not go through the gate’ is the same as the opposite (negation) of ‘You can go through the gate’ and, remembering gate is open = P,
- Q gate is closed = P ¯ + Q ¯ }+ }} you should be able to see that NOT = P ¯ + Q ¯ }+ }} or P,
Q ¯ }} = P ¯ + Q ¯ }+ }}
Example: Simplifying boolean equations using boolean algebra Simplify the following: ( A, B ¯ ) + A ¯ ¯ })+ }}}}
From looking at the truth table we can see that it equates to A,B, But we should also know how to get to this result by using boolean identities. Let’s give it a go: Using De Morgans Law: P + Q ¯ = P ¯, Q ¯ }= }. }}, Where P = ( A, B ¯ ) })} and Q = ( A ¯ ) })} Take each side separately and applying De Morgans Law convert the centre gate to an AND: ( A, B ¯ ) + ( A ¯ ) ¯ = ( A, B ¯ ¯ ), ( A ¯ ¯ ) })+( })}}=( }}}).( }})} Now dealing with the left hand side of our new equation ( A, B ¯ ¯ }}}} ), apply De Morgans Law again ( P, Q ¯ == P ¯ + Q ¯ }== }+ }} ) and cancel out the double bars: ( A ¯ + B ¯ ¯ ¯ ¯ ), ( A ) = ( A ¯ + B ), ( A ) }+ }}}}}).( )=( }+ ).( )} Multiply out both sides: ( A ¯, A ) + ( B, A ) }.A)+(B.A)} From the Identity A ¯, A = 0 }.A=0} we can replace the left hand side: 0 + ( B, A ) From the Identity 0 + X = X we can ignore the 0 leaving us with: B, A From the Identity X, Y = Y, X we can swap the values around: A, B = the value we calculated by truth table Let’s try another |
table>
Using Demorgans rule that: A, B ¯ = A ¯ + B ¯ }= }+ }} Making A, B ¯ + A = A ¯ + B ¯ + A }+A= }+ }+A} Using the boolean identity that A ¯ + A = 1 }+A=1} Making A ¯ + B ¯ + A = B ¯ + 1 }+ }+A= }+1} Using the boolean identity that B ¯ + 1 = 1 }+1=1} We simplify down to A, B ¯ + A == 1 }+A==1}
( A + B ¯ ), A ¯ ¯ }). }}}} Answer:
| A | B | A ¯ }} | B ¯ }} | A + B ¯ }} | ( A + B ¯ ), A ¯ }). }} | ( A + B ¯ ), A ¯ ¯ }). }}}} |
|---|---|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 1 | 1 | 0 |
| 0 | 1 | 1 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 | 1 |
| 1 | 1 | 0 | 0 | 1 | 0 | 1 |
If you’re catching on to this, you’ll notice that this is the equivalent of A + B, But we better check it with boolean algebra identities and De Morgans Law to confirm we have the correct answer.
Using De Morgans Law: P, Q ¯ = P ¯ + Q ¯ }= }+ }} Take each side separately and applying De Morgans Law convert the centre gate to an AND: ( A + B ¯ ), ( A ¯ ) ¯ = ( A + B ¯ ¯ ) + ( A ¯ ¯ ) }).( })}}= }}})+( }})}} Cancel out the double bars: ( A + B ¯ ¯ ) + ( A ¯ ¯ ) = ( A + B ¯ ¯ ) + ( A ) }}})+( }})}=( }}})+( )} Now dealing with the left hand side of our new equation, apply De Morgans Law again and cancel out the double bars: ( A ¯, B ¯ ¯ ¯ ¯ ) + ( A ) = ( A ¯, B ) + ( A ) }. }}}}})+( )=( }.)+( )} Multiply out both sides: ( A ¯ + A ), ( B + A ) }+A).(B+A)} From the Identity A ¯ + A = 1 }+A=1} we can replace the left hand side: 1. ( B + A ) From the Identity 1. X = X we can ignore the 1 leaving us with: B + A From the Identity X + Y = Y + X we can swap the values around: A + B = the value we calculated by truth table
( A + B ¯ ) + B ¯ })+B}}} Answer:
| A | B | A + B | A + B ¯ }} | ( A + B ¯ ) + B })+B} | ( A + B ¯ ) + B ¯ })+B}}} |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 |
| 1 | 1 | 1 | 0 | 1 | 0 |
A + B ¯ ) + B ¯ })+B}}} If you’re catching on to this, you’ll notice that this is the equivalent of A, B ¯ }}, But we better check it with boolean algebra identities and De Morgans Law to confirm we have the correct answer.
Using De Morgans Law: P, Q ¯ = P ¯ + Q ¯ }= }+ }} Take each side separately (P= A + B ¯ }} and Q= B ) and applying De Morgans Law convert the centre gate to an AND: ( A + B ¯ ) + B ¯ = ( A + B ¯ ¯ ), B ¯ })+B}}=( }}). }} Cancel out the double bars: ( A + B ¯ ¯ ), B ¯ = ( A + B ), B ¯ }}). }=(A+B). }} Multiply out both sides: ( A, B ¯ ) + ( B, B ¯ ) })+(B. })} From the Identity ( B, B ¯ ) = 0 })=0} we can replace the right hand side: ( A, B ¯ ) + 0 })+0} From the Identity X + 0 = X we can ignore the 0 leaving us with: A, B ¯ }} = the value we calculated by truth table
Fundamental Hardware Elements of Computers: De Morgan’s Laws – Wikibooks, open books for an open world
What are the two De Morgan’s theorem?
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.