Sorry to belabor this discussion. But I think I finally have a handle on it.Take a look at this "not" example show here:

http://hodoku.sourceforge.net/en/tech_ur.php There is possible 9 candidate occurring in the puzzle that somehow means the logic doesn't work. And then I read this in one of the links I posted above:

" A Bivalue Universal Grave or BUG is a state of the grid in which every unsolved cell has 2 candidates, without the presence of Hidden Singles."

[

http://hodoku.sourceforge.net/en/tech_ur.php ] Wiki Link 1

And then I read this text for "proof" of how this works:

" In a BUG, each unsolved cell has two candidates, and in each row, column and box constraint, each candidate value occurs exactly twice. If we can find a valid solution using half the number of remaining candidates, all the eliminated candidates would also satisfy each constraint exactly once. This other half will also give us a solution to the puzzle. There is also the possibility that the puzzle does not have a solution at all. In that case, the puzzle is also invalid. The existence of the BUG pattern allows us to conclude that the puzzle is invalid. It has either no solution or 2 solutions. "

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http://sudopedia.enjoysudoku.com/Bivalu ... Grave.html ] Wiki Link 2

Again, I hate to belabor point but consider my first first example:

https://imgur.com/I70Pv0MCell R6C3 has three possibilities (1, 7, 8). This is what got me hung up. It seemed to me 7 or 8 could solve the puzzle. I did not see why either the 7 or 8 could not solve the puzzle. So I tried the value of 8:

https://imgur.com/Rt961P9The value of 8 resulted in a contradiction in the puzzle as expected.

BUT, the value 7 in R6C3 solved the puzzle:https://imgur.com/7ipONK4And the value of 1 in cell R6C3 solved the puzzle:https://imgur.com/ZXEZf1nSo my initial problem in understanding this is I did not see why having a 7 or 8 in the cell could not be a possible solution to the puzzle. So I think the position taken with this technique is even though the 7 worked, just assume both will result in failure and pick the +1 candidate.

I think the must have condition for the Bug+1 to work is no other candidate in the entire puzzle can have more than 2 possible candidates in any house except the Bug+1 candidate which has 3 in the intersecting shared houses with the pivot cell.

Even though the 7 resulted in a positive result I think I now am on-board with the Bug+1. It's not too hard to find once you accept the conditions as I see them although I may be wrong. But I think I am right on this.

Here's an example where choosing either one of the non Bug+1 candidates resulted in a contradiction:

https://imgur.com/ooYiDO0