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Archangel Macsika

Let R₁ and R₂ be equivalence relation on X. Show that R₁ R₂ is an equivalence relation on X.

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We suppose thar R_1R_2\equiv R_1\cap R_2 because for R_1\circ R_2 statement is not valid.

1) reflexivity

Let x be any element in X, then <x,x>\in R_1,<x,x>\in R_2

because R1,R2 both are reflexive therefore by definition we have that <x,x>\in R_1\cap R_2=R_1R_2 . So reflexivity is proved.

2) symmetry

Let <x,y>\in R_1\cap R_2 . Then <x,y>\in R_1 and <x,y>\in R_2

Therefore <y,x>\in R_1 and <y,x>\in R_2 because both of R1,R2 are symmetrc as equivalences. By definition of intersection we have <y,x>\in R_1\cap R_2 therefore R1Ris symmetrical

3) transitivity

Let <x,y>,<y,z>\in R_1\cap R_2 . Therefore <x,y>,<y,z>\in R_1 and <x,y>,<y,z>\in R_2 . This this entails <x,z>\in R_1 and <x,z>\in R_2 because R1,R2 are transitive. So <x,z>\in R_1R_2 , therefore R1R2 is transitive.

In this way three basical properties of equivalence are proved and so R1R2 is equivalence.

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