What follows is, as far as I can tell, totally standard folklore. I have one particular point of confusion, other than that, I wanted to confirm that I am uttering the incantations correctly.

The smooth 4-dimensional Poincare conjecture (SPC4) is the statement that any smooth 4-dimensional manifold $\Sigma$ that is homeomorphic to $S^4$ is diffeomorphic to $S^4$. Is SPC4 equivalent to the conjecture that any smooth 4-manifold $B$ with $\partial B = S^3$ that is homotopy equivalent to $B^4$ is diffeomorphic to $B^4$?

**Going one direction:** Assume SPC4 is true.

If $B$ is a homotopy 4-sphere with $\partial B = S^3$ then we can fill the boundary with $B^4$ and we obtain a simply connected (by van Kampen) homology $S^4$ (by Mayer-Vietoris), which (by Hurewicz and Whitehead) must be a homotopy $S^4$, which (by Freedman) must be homeomorphic to $S^4$, which (assuming SPC4) is then diffeomorphic to $S^4$.

I imagine that we are about ready to conclude that $B$ must be standard - since it is now sitting inside of $S^4$ with its boundary bounding a standard $B^4$ on the other side. How do we finish up? Can we ambiently isotope $S^4$ so as to have the image of the $S^3$ be just the usual $S^3 \subset S^4$?

**Going the other direction:** Assume that every smooth homotopy $B^4$ with boundary $S^3$ is diffeomorphic to $B^4$.

Suppose that $\Sigma$ is a smooth homotopy $S^4$. Take a small 4-ball $B^4$ in $\Sigma$ and remove it. What is left (by van Kampen) is a smooth homotopy $B^4$ with boundary $S^3$ which (by assumption) is then diffeomorphic to $B^4$. Now (by Cerf) since $\Sigma$ is just the union of two copies of $B^4$, it is in fact diffeomorphic to $S^4$.