A knot $K$ is said to have Property P if every nontrivial Dehn surgery on $K$ yields a 3-manifold that is not simply connected. It is known that every knot except the unknot has Property P. I am wondering what can be said about a link that admits a nontrivial Dehn surgery that yields $S^3$.
An $n$-component link $L$ is said to have Property R if some Dehn surgery on $L$ yields $\sharp^n S^1 \times S^2$. The generalized Property R conjecture says that any such link, together with the framing used to obtain $\sharp^n S^1 \times S^2$ must be handleslide equivalent to an unlink with each component having framing 0. This conjecture is true for $n=1$ but unknown even for $n=2$ - see here. Note that by Kirby's Theorem, any two framed links that describe $\sharp^n S^1 \times S^2$ must differ by handleslides together with blowups and blowdowns - generalized Property R is asserting that the latter moves are not necessary in the case of $\sharp^n S^1 \times S^2$ for any pair of $n$-component descriptions.
Is there any sort of generalized Property P conjecture? There are certainly lots of links that that have a surgery that yields $S^3$ - for example any handlebody diagram for a 4-manifold without 1- or 3-handles. In fact, there is a conjecture that any simply connected smooth closed 4-manifold admits such a handlebody description (note: this implies S4PC) - if this is true then any such 4-manifold would yield such a framed link.
By considering the Hopf link with either $(0,0)$-framing or $(0,1)$-framing, we obtain two descriptions of $S^3$ that certainly are not handleslide equivalent - so Property P does not generalize naively like Property R. Maybe there is a bound $f(n)$, such that any two $n$-component framed link descriptions of $S^3$ require at most $f(n)$ blowups and blowdowns, together with handleslides in order to get between them?