State and prove De Morgan’s theorem.
Answer
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Hint: De Morgan’s law is a mathematical logic. Which is used in proposition logics and Boolean algebra, to denote the proposition of truth where $0$ is false and$1$ is true. Using $\overline{A+B}=\overline{A}\cdot \overline{B}$ and $\overline{A\cdot B}=\overline{A}+\overline{B}$ Or $\overline{A\cap B}=\overline{A}\cup \overline{B}$ and $\overline{A\cup B}=\overline{A}\cap\overline{B}$
Formula used:
$\overline{A+B}=\overline{A}\cdot \overline{B}$ or $\overline{A\cap B}=\overline{A}\cup \overline{B}$
$\overline{A\cdot B}=\overline{A}+\overline{B}$ or $\overline{A\cup B}=\overline{A}\cap\overline{B}$
Complete step by step solution:
De Morgan’s law is a mathematical logic. Which is used in proposition logics and Boolean algebra, to denote the proposition of truth where $0$ is false and$1$ is true. In English, the rules are expressed as negations, of conjunction and disjunction.
There are two De Morgan’s theorem :
De Morgan’s first theorem: the complement of a logical sum of two or more variables is equal to the logical product of the complement of the variables. $\overline{A+B}=\overline{A}\cdot \overline{B}$
In English, it is expressed as the negation of a disjunction is the conjunction of the negations. $\overline{A\cap B}=\overline{A}\cup \overline{B}$
If $A=0$ and $B=1$, then
$RHS: \overline{0+1}=\overline{1}=0$
$LHS:\overline{0}\cdot \overline{1}=1\cdot 0=0$
$RHS=LHS$
De Morgan’s second theorem: the complement of a logical product of two or more variables is equal to the logical sum of the complement of the variables.$\overline{A\cdot B}=\overline{A}+\overline{B}$
In English, it is expressed as the negation of a conjunction is the disjunction of the negation. $\overline{A\cup B}=\overline{A}\cap\overline{B}$
If $A=0$ and $B=1$, then
$RHS: \overline{0\cdot 1}=\overline{0}=1$
$LHS:\overline{0}+\overline{1}=1\cdot 0=1$
$RHS=LHS$
Note: It is worth remembering the law and its expression. It is used in truth tables and logic gates. We can see symmetry in both the laws, where complement of the product is the sum of individual complements, whereas complement of sum is the product of individual complements. In English, the rules are expressed as negations, of conjunction and disjunction.
Formula used:
$\overline{A+B}=\overline{A}\cdot \overline{B}$ or $\overline{A\cap B}=\overline{A}\cup \overline{B}$
$\overline{A\cdot B}=\overline{A}+\overline{B}$ or $\overline{A\cup B}=\overline{A}\cap\overline{B}$
Complete step by step solution:
De Morgan’s law is a mathematical logic. Which is used in proposition logics and Boolean algebra, to denote the proposition of truth where $0$ is false and$1$ is true. In English, the rules are expressed as negations, of conjunction and disjunction.
There are two De Morgan’s theorem :
De Morgan’s first theorem: the complement of a logical sum of two or more variables is equal to the logical product of the complement of the variables. $\overline{A+B}=\overline{A}\cdot \overline{B}$
In English, it is expressed as the negation of a disjunction is the conjunction of the negations. $\overline{A\cap B}=\overline{A}\cup \overline{B}$
If $A=0$ and $B=1$, then
$RHS: \overline{0+1}=\overline{1}=0$
$LHS:\overline{0}\cdot \overline{1}=1\cdot 0=0$
$RHS=LHS$
De Morgan’s second theorem: the complement of a logical product of two or more variables is equal to the logical sum of the complement of the variables.$\overline{A\cdot B}=\overline{A}+\overline{B}$
In English, it is expressed as the negation of a conjunction is the disjunction of the negation. $\overline{A\cup B}=\overline{A}\cap\overline{B}$
If $A=0$ and $B=1$, then
$RHS: \overline{0\cdot 1}=\overline{0}=1$
$LHS:\overline{0}+\overline{1}=1\cdot 0=1$
$RHS=LHS$
Note: It is worth remembering the law and its expression. It is used in truth tables and logic gates. We can see symmetry in both the laws, where complement of the product is the sum of individual complements, whereas complement of sum is the product of individual complements. In English, the rules are expressed as negations, of conjunction and disjunction.
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