For ease of reading, tel me recall RedEd's proof, that was in my first post. All the further posts from believers in this proof have only consisted of repeating it with no new argument. As a result, it remains a good starting point.
Red Ed wrote:Definition: an a/b/b/a pattern in a solution grid is anything isomorphic to that shown below:
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. . . | .
a . . | b
b . . | a
---------+---
. . . | .
Fact: if a solution grid (not necessarily unique) contains an a/b/b/a pattern on four unclued cells, C, then C=b/a/a/b is also a solution.
Theorem: if a puzzle-in-progress (that does not necessarily have a unique solution) has pencilmarks as shown below on four unclued cells then the bottom right value resolves to '3':
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. . . | .
1 . . | 2
2 . . | 13
---------+---
. . . | .
Proof: suppose to the contrary the bottom right value resolves to '1'. Then (vacuously) the solution grid contains the 1/2/2/1 pattern on four unclued cells, C. So, by the Fact above, C=2/1/1/2 is also a solution. But wait! - the pencilmarks do not allow that other solution - contradiction
The flaw I mentioned in this thread is related to using the condition 'unclued". Although it relies only on very basic logic, it seems to be too technical for being understood on this forum.
One more objection I now have is about the conclusion of the "Theorem" part. RedEd's conclusion is merely "contradiction". But wait! When a contradiction is found at the end of some proof, it only means that the supposed conclusion of the proof is inconsistent with all the hypotheses used for the proof.
What REdEd's proof really implies is:
- EITHER the value in the 4th cell must be 1 in the conditions of the theorem
- OR the conditions of the theorem are impossible (i.e. the a/b/b/ac pattern cannot appear if the 4 UR cells were unclued)
[Edited 2020/02/24 for clarification of my objection]: Of course, the second case is included in the first, but in standard mathematical practice, no one would merely jump to the general conclusion (first case above) when there are strong indications that the second particular case is indeed the real one.
Notice that if the second conclusion is true, this makes the theorem of the form FALSE => elimination. The theorem cannot even be said to be false (based only on this objection). It's merely an irrelevant tautology. So, this objection is not about the proof being correct or not, it's about it being irrelevant.
Note that both I and champagne have already come to the intuitive conclusion that the conditions are very unlikely to happen, based on vain tries for building an example.
[Edited 2020/02/24 for clarification of the remaining challenge]:It seems to me, as I already stated, that it should be a priority for the believers in the UR1.1 rule (with the "unclued" cells condition) to exhibit an example for a multi-solution puzzle (so that the implicit use of some assumption of uniqueness can be discarded Good luck!