Let me know if one thinks this is preferable to less depth, more steps. Hans, a fellow Ohioan, inspired the following alternative proof. Depth 20 I cannot 'see' upon puzzle inspection. My 'move horizon' is probably only about 11 (sometimes a tad more). Sometimes, I think that I prefer the longer chain versus the many shorter ones - If it is something I can 'see'. I have asked this question of others here before, and the preference they indicated was many short chains, so generally I present proofs that have many short chains versus one or two very long chains - for a puzzle such as this one. There is no 'true' way that I know of to assess which way is better: one long chain - or many shorter ones. I strive to reduce the depth of my proofs, but it does sometimes add signifcantly to the total number of steps.
Most puzzles can substantially unlock with one well chosen such proof by contradiction. Often, in terms of time spent to solve a puzzle, the strategy you proposed is more time efficient - in my opinion. I tried to do that with your idea, and came up with 20 sets - thus depth 20 since all those sets used in that one step. Adjust that for strong sets if you decided cellX = z because only candidate z left in row, column, box. If you count only the cells you needed to consider to reach the conclusion, it will be a close indication of depth. The route I found that was quickest to contradiction had depth 20: Meaning that I had to consider 20 strong sets to reach the contradiction.
Hi hans - There are many ways to go after the choice: What happens if a5=4.