GO Classes CS Test Series 2025 | Discrete Mathematics | Topic Wise Test 1 | Question: 2

Answer 1

$\color{red}{\text{Video Solution:}}$ https://www.youtube.com/watch?v=jCVNVv3F5u0&ab_channel=GOClassesforGATECS

For $n = 2 : \text{A}_{2} = p \rightarrow q ,$ which is a contingency.

For $n = 3 : \text{A}_{3} = p \rightarrow( q \rightarrow p ) ,$ which is a tautology.

For $n = 4 : \text{A}_{4} = p \rightarrow( q \rightarrow (p \rightarrow q) ) ,$ which is a tautology.

So, for $n =2, \text{A}$ is contingency, and for any value $n>2, \text{A}$ is tautology.

$\color{red}{\text{Video Solution:}}$ https://www.youtube.com/watch?v=jCVNVv3F5u0&ab_channel=GOClassesforGATECS

Answer 2

When n=2 :- A$_{2}$ = (p $\rightarrow$ q) $\equiv$ (p' + q)  => A$_{2}$ is contingency.

When n=3 :- A$_{3}$ = (p $\rightarrow$ (q $\rightarrow$ p)) $\equiv$ (p' + q' + p) $\equiv$ T  => A$_{3}$ is tautology.

When n=4 :- A$_{4}$ = (p $\rightarrow$ (q $\rightarrow$ (p $\rightarrow$ q))) $\equiv$ (p' + q' + p' + q) $\equiv$ T  => A$_{4}$ is tautology.

So, when n>2, the pattern for A$_{n}$ is (p' + q' + p' + q' + ... + q' + p) or (p' + q' + p' + q' + ... + p' + q), and both patterns are always true.

Therefore, when n>2, A$_{n}$ is tautology.