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 alongside what is typed below. What is typed below is an English example of what the argument is trying to prove.
Everyone loves a lover
The Baron loves Alma
Therefore Alma loves the Baron$\begin{split}&\forall x~(\exists y~\operatorname{Loves}(x,y)\to\forall z~\operatorname{Loves}(z,x))\\& \operatorname{Loves}(b,a)\\\hline\therefore~&\operatorname{Loves}(a,b)\end{split}$
This image contains the argument I need help proving in fitch notation. I can't figure out how to type out the quantifiers.
From $$\tag2 \operatorname{Loves}(b,a)$$ introduce $\exists$ to find $$\tag3\exists y\operatorname{Loves}(b,y)$$ Specialize $$\tag1\forall x(\exists y\operatorname{Loves}(x,y)\to \forall z\operatorname{Loves}(z,x)$$ to $$\tag4\exists y\operatorname{Loves}(b,y)\to \forall z\operatorname{Loves}(z,b)$$ Apply modus ponens to $(4)$ and $(3)$ to find $$\forall z\operatorname{Loves}(z,b)$$ and specialize to $$\operatorname{Loves}(a,b)$$