If is an equivalent relation then it is true that
supose p is an equivalent relation
(x,y) E p --> [x]p = [y]p
can anyone pls tell me whether the below approach is correct..
suppose (x,y) E p
let z be arbitrary and suppose
z E [x]p <--> (x,z) E p
like wise by using biconditional arrow, then u dont have to split it into 2 parts,
because i dont see any wrong in it.( since p is equivalent)
thanks in advance..