Question
Question: The negation of Boolean expression \(\sim s\vee (\sim r\wedge s)\) is equivalent to A) \(r\) B) ...
The negation of Boolean expression ∼s∨(∼r∧s) is equivalent to
A) r
B) s∧r
C) s∨r
D) ∼s∧∼r
Solution
We are asked to find the negation of ∼s∨(∼r∧s), we can write it as ∼(∼s∨(∼r∧s)). Simplify it and use the properties of Boolean algebra. After that, go on simplifying. Try it, you will get the answer.
Complete step by step solution:
We are asked to find the negation of ∼s∨(∼r∧s), we can write it as ∼(∼s∨(∼r∧s)).
Now simplifying that is solving the bracket in step by step way.
First let us simplify the negation.
∼(∼s∨(∼r∧s))=∼(∼s)∧(∼(∼r∧s))
Again, simplifying we get,
We know that, ∼(∼s)=s, so we get,
∼(∼s∨(∼r∧s))=s∧(∼(∼r∧s))
Now solving the bracket and we know ∼∧=∨.
Using above we get,
∼(∼s∨(∼r∧s))=s∧(∼(∼r)∨∼s))
On simplification we get,
∼(∼s∨(∼r∧s))=s∧(r∨∼s))
Now again simplifying we get,
∼(∼s∨(∼r∧s))=(s∧r)∨(s∧∼s)
Also, we know that, a∧∼a=F, we get,
∼(∼s∨(∼r∧s))=(s∧r)∨F
Also, a∨F=a, we get,
∼(∼s∨(∼r∧s))=(s∧r)
Therefore, the negation of Boolean expression ∼s∨(∼r∧s) is equivalent to (s∧r).
The correct answer is option (B).
Additional information:
Boolean algebra is used to analyze and simplify digital circuits. It is also called Binary Algebra or logical Algebra. The important operations performed in Boolean algebra are – conjunction (∧), disjunction (∨) and negation (¬). Any binary operation which satisfies the following expression is referred to as a commutative operation. Commutative law states that changing the sequence of the variables does not have any effect on the output of a logic circuit. Logic gates are an important concept if you are studying electronics. These are important digital devices that are mainly based on the Boolean function. Logic gates are used to carry out logical operations on single or multiple binary inputs and give one binary output.
Note:
Boolean algebra is the category of algebra in which the variable’s values are the truth values, true and false, ordinarily denoted 1 and 0 respectively. Here we have used some properties to get through with that. Keep in mind that they have asked for negation of ∼s∨(∼r∧s), do not start solving while looking at just the equation and read the problem thoroughly.