Exercise 3.1.11 (Replacement implies specification)

Proof. We have

where $Q(y,y):=P(y)$. In other words, we have defined the set in the axiom of specification using the notation of the axiom of replacement, meaning that the latter implies the former. □

Proof. Let $A$ be a set, and $P(x)$ a property. Consider the proposition $Q(x,y)$ defined by

Then for every $x\in A$, there is at most one $y$ for which $Q(x,y)$ is true, namely $x$ if $P(x)$ is true (otherwise there is no such $y$).

The axiom of replacement asserts that there is some set $B$ such that, for all object $y$,

Then

This shows that there is a set $B$ such that $y\in B$ if and only if $y\in A$ and $P(y)$: this is the axiom of specification. □