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I'm trying to understand the axiom of reeplacement in set theory. To my understanding, please tell me if I'm wrong, if there is a logical proposition $\varphi(x,y,w_{1},...,w_{n})$ and an arbitrary set $A$, then we can have the set $B=\{y:\exists x\in A(\varphi(x,y,w_{1},...,w_{n}))\}$.

Now, HERE is my problem. The formulation of the axiom in terms of logic makes explicit the fact that $\exists!y(x,y,w_{1},...,w_{n})$. This means then that the logical proposition needs to be a function. So basically I'm asking for examples where logical propositions are not functions. (To make this clearer, there is something I'm missing because I think that all proposition ar functions. For example the statemet $\forall x(x<y)\wedge (y+z=10)$ is a function of the variables $y$ and $z$, etc.)

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How is the statement that you put in your question a function? –  Apostolos Jul 16 '13 at 16:26
    
@Apostolos Because for every $y$ and $z$ there is only one value that the function $\varphi(y,z)=\forall x(x<y)\wedge(y+z=10)$ assigns to them. –  Daniela Diaz Jul 16 '13 at 16:30
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When we say that a formula $\varphi(x,y)$ is functional we mean that for every $x$ there is a unique $y$ such that $\varphi(x,y)$ is true. That is $x$ is the argument and $y$ is the output of the function. For example the formula $y=x+1$ has this property, because the successor of $x$ is unique. On the other hand the formula $(x=x\land y=y)$ is not functional. Because given a fixed $x$ there are more than one $y$ (in fact all $y$) that satisfy $\varphi(x,y)$. –  Apostolos Jul 16 '13 at 16:34
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Consider $\varphi(x,y)= y\in x$. This is not a function because $x=\{\varnothing,\{\varnothing\}\}$ does not have a unique $y$ satisfying this formula with $x$.

In fact, unless $A$ is a set of singletons, $\varphi(x,y)$ will not define a function on $A$.

Here is an example of why we must require that $\varphi$ is a function (after fixing the parameters) on the set $A$. Consider $A=\{\varnothing\}$ and $\psi(x,y)$ stating that $x\subseteq y$, formally: $$\psi(x,y)=\forall z(z\in x\rightarrow z\in y)$$

Now the collection $\{y\mid\exists x\in A.\psi(x,y)\}=\{y\mid y=y\}$, every set is a superset of the empty set. So this would be a proper class, which we already know is not a set. The axiom of replacement, as Hagen says, is telling us that if we can "uniformly rename all the elements of $A$" then the result is a set.

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In your proposed formulation of Replacement, let $A$ be any nonempty set and $\phi(x,y)\equiv x=x$. Then your $B$ would be tha all-class, which is not a set.

As you want to replace each element of $A$ with a single(!) new object $y$, you need the restriction mentioned, namely that (dropping the parameters $w_1,\ldots,w_m$ for simplicity) for all $x\in A$ there exists exactly* one object $y$ such that $\phi(x,y)$ holds. That is we can view $\phi$ not as a function from Sets$\times$Sets to $\{T,F\}$, but rather we have a function $$ x\mapsto \text{the unique }y\text{ such that }\phi(x,y)$$ * One could also work with at most one

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