blue wrote :

Yes it's true, as long as you're careful about (sic) what is meant by "the result in X is the not the same".

It might be better to think of it as that "the results in X are incompatible".

That would cover cases where:

1) one option forces some value into X, and the other option forces a different value into X.

2) one option forces X to be 'd', and the other eliminates 'd' from X.

3) neither option forces a value into X, but each candidate in X is eliminated for at least one of the options

I think bat is right after all.

Paraphrasing 2) : A <> 4 => X = d; B <> 4 => X <> d. So A <> 4 and B <> 4 => X = d and not d. Contradiction => one of A and B must be 4.

Paraphrasing 3) : A <> 4 => via, say, 2 AIC's => X <> a,b; B <> 4 => via, say, 3 AIC's => X <> c,d,e; If X = abcde then A <> 4 and B <> 4 => X is empty. Contradiction => one of A and B must be 4.

So 2) and 3) are proofs by contradiction. Shock, horror, howl of indignation

.

Paraphrasing 1) : A <> 4 ... some AIC 1 ... X = a; B <> 4 ... some other AIC 2 ... X = b. Rearranging we can say:

A <> 4 ... AIC 1 ... X = (a - b) ... reverse AIC 2 ... B = 4; and

B <> 4 ... AIC 2 ... X = (b - a) ... reverse AIC 1 ... A = 4. So shortening the notation:

A <> 4 ... AIC 1/2 ... B = 4; and

B <> 4 ... AIC 2/1 ... A = 4.

So 1) => A = 4 or B = 4, although I wouldn't say this means that the values that are forced into X (a and b) are incompatible. I think 1) was basically what bat was describing.

So what bat was describing, in a somewhat convoluted way, was simply a Type 1 AIC (ie AIC 1/2).

Of course this has nothing to do with Almost Locked Sets, and A and B don't have to share a house.

Leren