
The contrapositive of \[p\to \left( \sim q\to \sim r \right)\] is
(a) \[\left( \sim q\wedge r \right)\to \sim p\]
(b) \[\left( q\wedge r \right)\to \sim p\]
(c) \[\left( q\wedge \sim r \right)\to \sim p\]
(d) None of these
Answer
578.4k+ views
Hint: For solving this problem we use a simple definition of discrete mathematics. Contrapositive is nothing but negating the given statement which is denoted by \['\sim '\] and the symbol \['\to '\] is used for conditional statements. We use simple transformations listed below to solve this problem
\[\begin{align}
& \sim \left( p\to q \right)=\sim q\to \sim p \\
& \sim q\to \sim p=q\vee \sim p \\
& \sim \left( q\vee p \right)=\sim q\wedge \sim p \\
& \sim \left( \sim p \right)=p \\
\end{align}\]
Here, the symbols \['\wedge ,\vee '\] are denoted for ‘and’, ‘or’ statements respectively.
Complete step by step answer:
Let us take the given statement as
\[x=p\to \left( \sim q\to \sim r \right)\]
We know that the contrapositive of a statement is nothing but applying the negation to the whole statement.
By applying the negation to given statement we will get
\[\Rightarrow x=\sim \left( p\to \left( \sim q\to \sim r \right) \right)\]
We know that the negation of a conditional statement is given as \[\sim \left( p\to q \right)=\sim q\to \sim p\].
By applying this formula to above equation we will get
\[\Rightarrow x=\sim \left( \sim q\to \sim r \right)\to \sim p\]
We know that the conditional statement of two negation statements is given as \[\sim q\to \sim p=q\vee \sim p\]
By applying this formula to above equation we will get
\[\Rightarrow x=\sim \left( q\vee \sim r \right)\to \sim p\]
By taking the negation of first statement in above equation inside then the ‘or’ symbol is converted to ‘and’ symbol negation is applied to both inside statements that is \[\sim \left( q\vee p \right)=\sim q\wedge \sim p\]
So, the above equation can be modified as
\[\Rightarrow x=\left( \sim q\wedge \sim \left( \sim r \right) \right)\to \sim p\]
We know that the negation of the negation statement given the original statement that is \[\sim \left( \sim p \right)=p\].
By applying this formula to above statement we will get
\[\Rightarrow x=\left( \sim q\wedge r \right)\to \sim p\]
Therefore, the answer is \[\left( \sim q\wedge r \right)\to \sim p\].
So, the correct answer is “Option a”.
Note: Students may make mistakes in applying the discrete mathematics formulas that are mentioned above. In some formulas there will be interchanging of positions of statements after applying the negation to conditional statements. That is
\[\sim \left( p\to q \right)=\sim q\to \sim p\]
Here, students may miss in interchanging and takes the formula as
\[\sim \left( p\to q \right)=\sim p\to \sim q\]
This will be wrong. So the application of formula needs to be taken care of.
\[\begin{align}
& \sim \left( p\to q \right)=\sim q\to \sim p \\
& \sim q\to \sim p=q\vee \sim p \\
& \sim \left( q\vee p \right)=\sim q\wedge \sim p \\
& \sim \left( \sim p \right)=p \\
\end{align}\]
Here, the symbols \['\wedge ,\vee '\] are denoted for ‘and’, ‘or’ statements respectively.
Complete step by step answer:
Let us take the given statement as
\[x=p\to \left( \sim q\to \sim r \right)\]
We know that the contrapositive of a statement is nothing but applying the negation to the whole statement.
By applying the negation to given statement we will get
\[\Rightarrow x=\sim \left( p\to \left( \sim q\to \sim r \right) \right)\]
We know that the negation of a conditional statement is given as \[\sim \left( p\to q \right)=\sim q\to \sim p\].
By applying this formula to above equation we will get
\[\Rightarrow x=\sim \left( \sim q\to \sim r \right)\to \sim p\]
We know that the conditional statement of two negation statements is given as \[\sim q\to \sim p=q\vee \sim p\]
By applying this formula to above equation we will get
\[\Rightarrow x=\sim \left( q\vee \sim r \right)\to \sim p\]
By taking the negation of first statement in above equation inside then the ‘or’ symbol is converted to ‘and’ symbol negation is applied to both inside statements that is \[\sim \left( q\vee p \right)=\sim q\wedge \sim p\]
So, the above equation can be modified as
\[\Rightarrow x=\left( \sim q\wedge \sim \left( \sim r \right) \right)\to \sim p\]
We know that the negation of the negation statement given the original statement that is \[\sim \left( \sim p \right)=p\].
By applying this formula to above statement we will get
\[\Rightarrow x=\left( \sim q\wedge r \right)\to \sim p\]
Therefore, the answer is \[\left( \sim q\wedge r \right)\to \sim p\].
So, the correct answer is “Option a”.
Note: Students may make mistakes in applying the discrete mathematics formulas that are mentioned above. In some formulas there will be interchanging of positions of statements after applying the negation to conditional statements. That is
\[\sim \left( p\to q \right)=\sim q\to \sim p\]
Here, students may miss in interchanging and takes the formula as
\[\sim \left( p\to q \right)=\sim p\to \sim q\]
This will be wrong. So the application of formula needs to be taken care of.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

