I am supposed to show that the axiom of replacement implies the axiom of seperation and I am not sure if my solution is correct.
I am trying to show that given any set A and Property P(x) there is a subset of A which contains exactly the Elements for which P(x) is correct.(set 1)
I choose the Property P(x,y)=P(x) and y=x. By the replacement axiom (y:P(x,y) is true for some x in A) (set 2) is a set.
I now have to show that set 1 = set 2. If x is an Element of set 1 than the Element y=x must be an Element of set 2 because for x in A we have y=x and P(x). If y is an Element of set 2 than by the definition of set 2 y=x and P(x) and x is in A so y is also an Element of set 1.
I am very new to proving stuff so I am grateful for any feedback.
Regards James
