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We suppose thar because for statement is not valid.
Let x be any element in X, then
because R1,R2 both are reflexive therefore by definition we have that . So reflexivity is proved.
Let . Then and
Therefore and because both of R1,R2 are symmetrc as equivalences. By definition of intersection we have therefore R1R2 is symmetrical
Let . Therefore and . This this entails and because R1,R2 are transitive. So , therefore R1R2 is transitive.
In this way three basical properties of equivalence are proved and so R1R2 is equivalence.