$Pa\rightarrow \exists y Qy \vdash \exists y (Pa\rightarrow Qy)$
My friend asked me to prove this using natural deduction. He knows I studied logic but I know little about natural deduction since I studied logic only by hilbert style.
Using inference rules he presented I thought a lot, but couldn't find any proof easier than these two. First, introduce $Pa\rightarrow \exists y Qy$ as a premise.
1-1. Assume $\neg\exists y(Pa\rightarrow Qy)$ and derive $\forall y\neg(Pa\rightarrow Qy)$
1-2. Derive $\neg(Pa\rightarrow Qb)$ and find subproof of $(Pa\wedge \neg Qb)$
1-3. Derive $\forall y\neg Qy$ from $\neg Qb$ and $\exists y Qy$ from $Pa$
1-4. Derive $\neg\exists yQy$ from $\forall y \neg Qy$ and conclude there's a contradiction
2-1. Drive $\neg Pa \vee\exists yQy$
2-2. Derive $Pa\rightarrow Qb$ each from $\neg Pa$ and $\exists y Qy$ and conclude $\exists y(Pa\rightarrow Qy)$
Since I haven't tried this in a full content, I don't know how long proof will be but I guess it will be.. I think there should be easier proof than these but I don't know how to find it.