if events X and Y are independent, then for intersection we multiply the probability
P(Y∩X) = P(Y) * P(X)
We know that

Now we replace P(Y) * P(X) for P(Y∩X)

Cancel out P(X)
So

Like that

Now we replace P(X) * P(Y) for P(X∩Y)

Cancel out P(Y)
So

P(Y | X) = P(Y) and P(X | Y) = P(X) are true