Step 1: Reflexive Property.
For the relation \( R \) to be reflexive, it must contain the pair \( (a,a) \) for every element \( a \in I \). Here, the relation does not contain \( (4,4), (5,5), (6,6), (7,7) \), so \( R \) is not reflexive.
Step 2: Symmetric Property.
For the relation to be symmetric, if \( (a,b) \) is in the relation, then \( (b,a) \) must also be in the relation. Here, \( (4,5) \) and \( (5,4) \) are present, and similarly for the other pairs, making \( R \) symmetric.
Step 3: Transitive Property.
For the relation to be transitive, if \( (a,b) \) and \( (b,c) \) are in the relation, then \( (a,c) \) must also be in the relation. However, \( (4,5) \) and \( (5,4) \) should imply \( (4,4) \), but \( (4,4) \) is not in the relation, so \( R \) is not transitive.
Step 4: Antisymmetric Property.
For the relation to be antisymmetric, if \( (a,b) \) and \( (b,a) \) are in the relation, then \( a = b \). In this case, both \( (4,5) \) and \( (5,4) \) are present, but \( 4 \neq 5 \) and \( 5 \neq 4 \), so \( R \) is not antisymmetric.
Step 5: Conclusion.
The relation does not have the transitive property, so the correct answer is (1) A, C and D only.
Find the least upper bound and greatest lower bound of \( S = \{X, Y, Z\} \) if they exist, of the poset whose Hasse diagram is shown below:
Suppose \( D_1 = (S_1, \Sigma, q_1, F_1, \delta_1) \) and \( D_2 = (S_2, \Sigma, q_2, F_2, \delta_2) \) are finite automata accepting languages \( L_1 \) and \( L_2 \), respectively. Then, which of the following languages will also be accepted by the finite automata:
(A) \( L_1 \cup L_2 \)
(B) \( L_1 \cap L_2 \)
(C) \( L_1 - L_2 \)
(D) \( L_2 - L_1 \)
Choose the correct answer from the options given below:
Match LIST-I with LIST-II \[\begin{array}{|c|c|c|}\hline \text{ } & \text{LIST-I} & \text{LIST-II} \\ \hline \text{A.} & \text{A Language L can be accepted by a Finite Automata, if and only if, the set of equivalence classes of $L$ is finite.} & \text{III. Myhill-Nerode Theorem} \\ \hline \text{B.} & \text{For every finite automaton M = $(Q, \Sigma, q_0, A, \delta)$, the language L(M) is regular.} & \text{II. Regular Expression Equivalence} \\ \hline \text{C.} & \text{Let, X and Y be two regular expressions over $\Sigma$. If X does not contain null, then the equation $R = Y + RX$ in R, has a unique solution (i.e. one and only one solution) given by $R = YX^*$.} & \text{I. Arden's Theorem} \\ \hline \text{D.} & \text{The regular expressions X and Y are equivalent if the corresponding finite automata are equivalent.} & \text{IV. Kleen's Theorem} \\ \hline \end{array}\]
\[\text{Matching List-I with List-II}\]
Choose the correct answer from the options given below: