I can't figure out how to prove this argument in fitch notation, I continue to run into issues in proving it. Please help me solve this and if you solve it please provide the step numbers with the rules.
Please see the link to the attached image, it contains the actual argument.
$$\begin{split}&\forall x~(P(x)\lor\exists y~R(x,y))\\&\forall x~(Q(x)\to\lnot P(x))\\&\forall x~\forall y~(R(x,y)\to S(y))\\\hline\therefore~&\exists x~Q(x)~\to~\exists y~S(y)\end{split}\qquad\raise{4ex}{\text{Notice that the conclusion}\\\text{ is a conditional statement}}$$
How exactly your proof needs to be will depend on the exact rules you will be using, in particular the existential quantifier might be tricky. A proof outline using the rules in the proof system at open logic project is as follows:
Since your desired conclusion is a conditional, start by assuming the hypothesis $\exists x Qx$. Introduce a subproof with hypothesis $Qa$. Apply $\forall $-elimination on the second assumption to yield $Qa\rightarrow \neg Pa$, which from the hypothesis and $\rightarrow$-elimination yields $\neg Pa$. Similarly, apply $\forall$-elimination on the first assumption to get $Pa \vee\exists yRay$. From $\neg Pa$, we can apply the disjunctive syllogism to get $\exists yRay$. Introduce another subproof with hypothesis $Rab$. Apply $\forall$-elimination twice on the third assumption to get $Rab\rightarrow Sb$, from which with $\rightarrow$-elimination and $Rab$ we have $Sb$. Now $\exists$-introduction on $Sb$ yields $\exists ySy$, and $\exists$-elimination on $\exists yRay$ lets us pull the conclusion $\exists ySy$ out of the $Rab$ subproof into the $Qa$ subproof. Almost there, we can apply $\exists$-elimination again to pull $\exists ySy$ out of the $Qa$ subproof into the $\exists xQx$ hypothesis, and finally $\rightarrow$-introduction lets us conclude $\exists xQx\rightarrow\exists ySy$.
I leave it to you to format it according to whatever standards you need to format it to.