What would the slice-ribbon conjecture imply for 4-dimensional topology?  

I've heard people speak of the slice-ribbon conjecture as an approach to the 4-dimensional smooth Poincare conjecture, and to the classification of homology 3-spheres which bound homology 4-balls. But I've never understood what they were talking about. 



What would the slice-ribbon conjecture imply?
I think of the ribbon-slice conjecture as a **wish** that would simplify certain 4D questions. Let me explain this in 3 examples.


1. Given an embedded "ribbon disk" in 4-space (where the Morse function has no local maxima) one can push it up into 3-space and obtain an immersed disk (whose boundary is still the given knot) where the singularities are mild: these are the so called "ribbon singularities", arcs of double points such that on one of the sheets, the arc lies in the interior. (Picture...) One would actually call this immersed disk in 3-space a "ribbon" (that is allowed to cut through itself). It contains the information about the embedded disk in 4-space by pushing one sheet of each ribbon singularity into the forth dimension. 
There is a fairly obvious algorithm how to create all such ribbons in 3-space, starting from the unlink and adding bands. 
No such simple 3D-picture exists for arbitrary slice disks and one may **wish** that any slice knot is ribbon.
2. A knot K is slice if and only if there is a ribbon knot R such that the connected sum K # R is ribbon. One may **wish** that one didn't have to stabilize.
3. Consider the monoid M of oriented knots under connected sum. If -K is the reversed mirror image of K then K # (-K) is ribbon. So it's very tempting to try to turn M into a group (where -K would become the inverse of K) by identifying two knots K' and K if K' # (-K) is ribbon. But the **wish** doesn't come true: this is not an equivalence relation and if we force it to be one then by 2 we end up with the knot concordance group (where two knots K' and K are identified if K' # (-K) is slice).


It is amazing that there are no proposed counter examples to this conjecture, not even for links.


