We suppose thar R1R2≡R1∩R2 because for R1∘R2 statement is not valid.
1) reflexivity
Let x be any element in X, then <x,x>∈R1,<x,x>∈R2
because R1,R2 both are reflexive therefore by definition we have that <x,x>∈R1∩R2=R1R2 . So reflexivity is proved.
2) symmetry
Let <x,y>∈R1∩R2 . Then <x,y>∈R1 and <x,y>∈R2
Therefore <y,x>∈R1 and <y,x>∈R2 because both of R1,R2 are symmetrc as equivalences. By definition of intersection we have <y,x>∈R1∩R2 therefore R1R2 is symmetrical
3) transitivity
Let <x,y>,<y,z>∈R1∩R2 . Therefore <x,y>,<y,z>∈R1 and <x,y>,<y,z>∈R2 . This this entails <x,z>∈R1 and <x,z>∈R2 because R1,R2 are transitive. So <x,z>∈R1R2 , therefore R1R2 is transitive.
In this way three basical properties of equivalence are proved and so R1R2 is equivalence.
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