Aran, I think you have misread my post on your first point. I have no hypothetical objections to your hypothetical object at all, in fact that you managed to compose it is admirable! That's why I built up my stripped down version of it.

But now I must apologise for mistakenly editing out two instances of (12) just as I was posting my piece, which I've now restored.

Repeated:

- Code: Select all
`|---------|-------|---------|`

| p1 . . | . . . | . . x1 | RCCs = 1 & 2

| pq . . | . . . | . . y12 |

| 6 . . | . . . | . . 7 | p = 3 and/or 4 (1234)ANS:r149c1

|---------|-------|---------| q = 5 and/or 7 (57+1234)AHS:r267c1

| p . . | . . . | . . 8 |

| 9 . . | . . . | . . xy | x = 4 and/or 5 (1245)ANS:r169c9

| pq . . | . . . | . . x | y = 6 and/or 9 (69+1245)AHS:r258c9

|---------|-------|---------|

| q12 . . | . . . | . . 3 |

| 8 . . | . . . | . . xy |

| p2 . . | . . . | . . x2 |

|---------|-------|---------|

(134=234)r1c9c1 – (245=145)r169c9 – Loop => r26c1> 34, r58c9 <> 45, r1c2345678 <> 1, r9c2345678 <> 2

My reasoning was that there can only be one of each in the ANSs because if there were more there would be alternative cells where they could be true and the weak links in rows 1 & 9 wouldn't necessarily follow (as your original description showed).

If the AHSs were extended to include some of the cells I've shown as solved, then (12) could be added to those as well, but they couldn't exist together in any other cells containing p in c1 or x in c9.

My conclusion was that although your hypothetical DL-ANSs could exist, there would generally be a variety of other ways the eliminations could be made.