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In either diagram of the Stevedore knot, the evident ribbon band can be cut to leave a two-component unlink. Cap those components by disjoint disks in $B^4$ and restore the band. This constructs a ribbon disk, hence proves that the Stevedore knot is slice.

The two displayed Stevedore diagrams give two distinct ribbon-band presentations. Use one below the equator and the reverse of the other above it. In banded-unlink diagram language, draw their common unlink and include the two ribbon bands, one from each presentation. Reading the movie from bottom to top gives the indicated birth level, the two saddle bands, and the death level. The resulting surface is knotted: the two-band movie is the standard presentation obtained by gluing the two Stevedore ribbon disks, and a van Kampen calculation retains a noncyclic quotient coming from the trefoil group, whereas the complement of the unknotted model has cyclic fundamental group.

There is a terminology issue in the question. If “2-knot” means an embedded $S^2$, as it normally does, one minimum, two index-one saddles, and one maximum have Euler characteristicso the resulting orientable surface knot is a torus, not a two-sphere. The construction above answers the question under the broader usage in which “2-knot” means a connected knotted surface. Under the standard narrow definition, the requested critical-point data are impossible.

Solved by gpt-5.6-sol high.