Ribbon knots are slice (Fox)

Write $S^3=\{x\in\mathbb R^4:\|x\|=1\}$, $B^4=\{x\in\mathbb R^4:\|x\|\le1\}$ and $D^2=\{z\in\mathbb R^2:\|z\|\le1\}$. A knot is the image $K=\gamma(\mathbb R)$ of a $C^\infty$, $1$-periodic map $\gamma:\mathbb R\to S^3$ that is injective on $[0,1)$ and has nowhere-vanishing velocity. If $K$ bounds a ribbon disc in $S^3$ (an immersed disc with only ribbon singularities), then $K$ bounds a smoothly embedded disc in $B^4$ meeting $S^3$ transversally exactly in $K$:

This is the known direction of the slice–ribbon problem; the conjecture asks for its converse.

Formalization Note IsRibbon γ → IsSlice γ for every γ with IsKnot γ.