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Axiom $3.6$ (Replacement). Let $A$ be a set. For any object $x \in A$ and any object $y$, suppose we have a statement $P(x, y)$ pertaining to $x$ and $y$, such that for each $x\in A$ there is at most one $y$ for which $P(x,y)$ is true. Then there exists a set $\{y: P(x, y) \text{ is true for some } x \in A\}$ such that for апy object $z$, $$ z\in \{y : P(x, y)\text{ is true for some } x \in A\} \iff P(x,z)\text{ is true for some } x \in A.$$

Axiom $3.5$ (Specification). Let $A$ be a set, and for each x$\in$ $A$, let $P(x)$ be a property pertaining to $x$ (i.e., $P(x)$ is either a true statement or a false statement). Then there exists a set, called $\{x \in A : P(x) \text{ is true}\}$ (or simply $\{x \in A : P(x) \text{ for short}\}$), whose elements are precisely the elements $x$ in $A$ for which $P(x)$ is true. In other words, for any object $y$, $$у \in \{x \in A: P(x)\text{ is true}\} \iff (y \in A \text{ and } P(y)\text{ is true}).$$

I have to show that $3.6\implies 3.5.$

Proof: By $(3.6)$ we can assume the following set $$\{x:P(x,x)\text{ is true for some }x\in A\}.$$ Let $Q(x)=P(x,x)$ then we get the set $$\{x\in A:Q(x) \text{ is true for some }x\},$$ which is what $(3.5)$ wants. Is this proof correct?

PS. I have read other answers to this question on MSE, but none of them use this formulation of the axiom and I guess use more formal notation. I am learning from Tao's Analysis book and so I've not been introduced to such notation.

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    The choice is tags is truly baffling. – 2017-01-21
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    You can see : [proving Separation from Replacement](http://math.stackexchange.com/questions/680376/proving-separation-from-replacement). – 2017-01-21

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Your argument isn't quite correct. But it's on the way.

You are given a property $P$ and you want to separate a subset of $A$ which is specified by the property $P$.

For this, you need to find a property $Q(x,y)$ for which if $x\in A$, then there is at most a single $y$ such that $Q(x,y)$ holds, and you need to choose $Q$ in such a way that you get exactly the subset of $A$ specified by $P$.

The obvious solution, of course, is $Q(x,y)$ to be $x=y\land P(x)$. I will leave you to prove that it satisfies the conditions needed to apply Replacement on $Q$, and of course $\{y\mid Q(x,y)\text{ holds for some }x\in A\}$ is exactly—by the definition of $Q$—the same $\{x\in A\mid x=x\land P(x)\}$ which in turn is exactly $\{x\in A\mid P(x)\}$.

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    Why are we starting with a property that only depends on $x$? Axiom 3.6 pertains to two variables $x$ and $y$. – 2018-05-28
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    You want to *use* 3.6 in order to *prove* 3.5. The statement of Specification is "If bla is given, then brau exists". So we start with bla, as it is given, and we use 3.6 to prove that brau exists. But in order to use 3.6, we again have a statement of the form "If brut is given, then blat exists". So from bla, we define brut, and from blat we derive brau. – 2018-05-28