Let R and S be two relations on a set A. Then | vedclass

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(D) $1$. If $R$ and $S$ are transitive,then for any $(x, y) \in R \cap S$ and $(y, z) \in R \cap S$,we have $(x, y) \in R, (y, z) \in R \implies (x, z

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All of these.

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(D) $1$. If $R$ and $S$ are transitive,then for any $(x, y) \in R \cap S$ and $(y, z) \in R \cap S$,we have $(x, y) \in R, (y, z) \in R \implies (x, z) \in R$ and $(x, y) \in S, (y, z) \in S \implies (x, z) \in S$. Thus $(x, z) \in R \cap S$,so $R \cap S$ is transitive.$2$. If $R$ and $S$ are reflexive,then for all $a \in A$,$(a, a) \in R$ and $(a, a) \in S$. Thus $(a, a) \in R \cap S$,so $R \cap S$ is reflexive.$3$. If $R$ and $S$ are symmetric,then for any $(x, y) \in R \cup S$,either $(x, y) \in R$ or $(x, y) \in S$. Since $R$ and $S$ are symmetric,$(y, x) \in R$ or $(y, x) \in S$,which means $(y, x) \in R \cup S$. Thus $R \cup S$ is symmetric.Therefore,all statements are correct.

Let $R$ and $S$ be two relations on a set $A$. Then which of the following statements is true?

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