A pivot cell with two candidates {X,Y} sees two wings {X,Z} and {Y,Z}; digit Z is eliminated from cells seeing both wings.
A "pivot" cell has candidates {X,Y}. It sees two "wing" cells: one with {X,Z} and one with {Y,Z}. If pivot takes X, the first wing must take Z. If pivot takes Y, the second wing takes Z. Either way, Z must be in one of the wings — so any cell seeing both wings can have Z eliminated.
This technique uses "pincer" logic and is very powerful.
79 | 2 | 1 | 4 | 57 | 8 | 3 | 9 | 8 |
| 4 | 6 | 5 | 124 | 2 | 3 | 1 | 4 | 6 |
| 2 | 9 | 4 | 579 | 8 | 6 | 5 | 138 | 7 |
| 6 | 158 | 279 | 3 | 4 | 1 | 459 | 2 | 8 |
37 | 469 | 168 | 249 | 35 | 268 | 149 | 57 | 6 |
| 9 | 145 | 8 | 157 | 267 | 9 | 358 | 6 | 7 |
| 1 | 5 | 357 | 6 | 9 | 5 | 679 | 8 | 2 |
| 5 | 389 | 6 | 279 | 1 | 458 | 239 | 5 | 3 |
| 3 | 7 | 5 | 348 | 7 | 2 | 4 | 3 | 1 |
Pivot R5C5={3,7}. Wing1 R5C1={3,9}, Wing2 R2C5={7,9}. → R2C1 sees both wings → Eliminate 9 from R2C1.
This 9×9 puzzle is solver-verified to require this technique on its solution path.