Formal Proof Fitch have been working on this proof and can't seem to figure out how to do a critical step. I am trying to derive the universal quantifier @x and @y. I have derived @z already. Any help is appreciated thanks]1

Here is the original question:

 A: $\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}$
You are eliminating the two existentials by assuming witness terms.
If you want to introduce three universals you must also assume three arbitrary terms.
$$\fitch{\exists x~\exists y~(C(x)\wedge C(y)\wedge x\neq y\wedge\forall z~(C(z)\to(z=x\vee z=y)))}{\fitch{[a]~\exists y~(C(a)\wedge C(y)\wedge a\neq y\wedge\forall z~(C(z)\to(z=a\vee z=y)))}{\fitch{[b]~C(a)\wedge C(b)\wedge a\neq b\wedge\forall z~(C(z)\to(z=a\vee z=b))}{\forall z~(C(z)\to(z=a\vee z=b))\\\fitch{[c]}{C(c)\to(c=a\vee c=b)\\\fitch{[d]}{C(d)\to(d=a\vee d=b)\\\fitch{[e]}{C(e)\to(e=a\vee e=b)\\\fitch{C(c)\wedge C(d)\wedge C(e)}{~\vdots~\text{some stuff happens here}\\~\vdots\\c=d\vee c=e\vee d=e}\\C(c)\wedge C(d)\wedge C(e)\to c=d\vee c=e\vee d=e}\\\forall z~(C(c)\wedge C(d)\wedge C(z)\to c=d\vee c=z\vee d=z)}\\\forall y~\forall z~(C(c)\wedge C(y)\wedge C(z)\to c=y\vee c=z\vee y=z)}\\\forall x~\forall y~\forall z~(C(x)\wedge C(y)\wedge C(z)\to x=y\vee x=z\vee y=z)}\\\forall x~\forall y~\forall z~(C(x)\wedge C(y)\wedge C(z)\to x=y\vee x=z\vee y=z)}\\\forall x~\forall y~\forall z~(C(x)\wedge C(y)\wedge C(z)\to x=y\vee x=z\vee y=z)}$$
