Proof of Composite Maps into Classes, Images of Singletons, and Agreement of Maps by Induction

Throughout, every element of a class is a set (Elements Are Sets, and the Empty Set and the Pair of Two Sets Exist and Are Unique (i)).

For (i): let $x\in X$. By Maps between Classes: Values, Pairs, Extensionality, Restrictions, Constant Maps and Preimages (i), $f(x)$ is a set in $A$ and $g(x)$ a set in $B$, so $(f(x),g(x))\in A\times B$ by Maps between Classes: Values, Pairs, Extensionality, Restrictions, Constant Maps and Preimages (viii), and then $W((f(x),g(x)))$ is a set in $C$, by Maps between Classes: Values, Pairs, Extensionality, Restrictions, Constant Maps and Preimages (i) applied to $W$. Thus the expression

$$t(x)=W((f(x),g(x))),$$

with parameters $W$, $f$ and $g$, satisfies $t(x)\in C$ for every $x\in X$ (its defined set symbols are used properly, by Maps between Classes: Values, Pairs, Extensionality, Restrictions, Constant Maps and Preimages (i) and Maps between Classes: Values, Pairs, Extensionality, Restrictions, Constant Maps and Preimages (viii)). Now Maps and Relations Given by Formulas (i), applied with $a=X$, $b=C$ and this $t$, gives a map $h:X\to C$ with $h(x)=t(x)$ for every $x\in X$.

For (ii): since $y$ is a set, The Image and the Preimage of a Class under a Class (i) and Class Abstraction: the Class of All Sets Satisfying a Predicative Formula (i) give that $y\in R[\{x\}]$ if and only if there is a set $u$ with $u\in\{x\}$ and $(u,y)\in R$. By The Empty Set, the Unordered Pair and the Singleton (iii), the only element of $\{x\}$ is $x$. So if $(x,y)\in R$, the set $u=x$ is such a set; conversely, such a set $u$ equals $x$, and then $(x,y)=(u,y)\in R$.

For (iii): let $\varphi$ be the condition on the set variable $k$, with parameters $T$, $S$ and $M$,

$$k\in T\Rightarrow\exists w\,\bigl((k,w)\in S\wedge(k,w)\in M\bigr)$$

(it is expressed by a predicative formula, namely by itself, since it quantifies over the set $w$ only).

We first show that, for every set $k$, $\varphi(k)$ holds if and only if $k\in T$ implies $S(k)=M(k)$. It suffices to show, for $k\in T$, that a set $w$ with $(k,w)\in S$ and $(k,w)\in M$ exists if and only if $S(k)=M(k)$. So let $k\in T$; then $k\in A$ and $k\in B$. If $S(k)=M(k)$, then $w=S(k)$ is a set with $(k,w)\in S$ and $(k,w)=(k,M(k))\in M$, by Maps between Classes: Values, Pairs, Extensionality, Restrictions, Constant Maps and Preimages (i). Conversely, if a set $w$ satisfies $(k,w)\in S$ and $(k,w)\in M$, then $w=S(k)$ and $w=M(k)$ by Maps between Classes: Values, Pairs, Extensionality, Restrictions, Constant Maps and Preimages (ii), so $S(k)=M(k)$.

This equivalence turns the first hypothesis into $\varphi(0)$ and the second into: $\varphi(n)$ implies $\varphi(n+1)$ for every $n\in\mathbb{N}_{0}$. By Induction on ℕ₀ from 0 for a Condition Expressed by a Predicative Formula (ii), $\varphi(k)$ holds for every $k\in\mathbb{N}_{0}$, which by the equivalence again says that $S(k)=M(k)$ for every $k\in\mathbb{N}_{0}$ with $k\in T$.