Dual AALS link

In this post I will showcase a rather interesting interaction between two AALS that share a cell. This came up within an SE 8.3 puzzle I was solving and delighted me with its simplicity and its practicality (it immediately reduced the puzzle to basics). It is very similar to Sue de Coq. I'm sure it's known but I've never seen it brought up before. Typically when chaining off AALS, the easiest option tends to be to use region-wise interactions with ALS weakly linking 2 candidates at once. 

Example puzzle

.4.1....7......2..7...4..1...6..5..83..9......5.6...7......3..68...1..2...1...39. - SE 8.3

After basics direct your attention to the bottom-left 2 boxes of the puzzle.

Do you see it? Let me colour in some cells.

We have two AALS:
AALS A = {2457}r7c24

AALS B = {2457}r78c4

These AALS intersect within the cell r7c4, and the non-intersecting cells share no digits. Also, both are DoF 2, and the candidates within the intersecting cell can be partitioned into 2 sets of 2 digits each, which are weakly linked to each other.

If all the blue candidates were removed from r7c4, the purple AALS would become a naked pair. And vice versa, if all the purple candidates were removed from r7c4, the blue AALS would become a naked pair. This gives us the AIC (45)r78c4 = (27-45)r7c4 = (27)r7c24.

This isn't enough to eliminate candidates but it can easily be extended to do so:

(45)r78c4 = (27-45)r7c4 = (27)r7c24 - (7)r7c3 = (7-4)r5c3 = (4)r5c6 => r4c4,r9c6<>4

STTE

Can be generalised to larger AALS comprised of more than 2 cells, even to AAALS or even higher DoF. This is hardly a discovery, but I'd never considered such an interaction before, and it should be useful to keep in mind when chaining in the future. 

Replacing the ALS with AHS may not always lead to the same eliminations, seeing as the {45} pair is contained in both a column and a box so there are 2 equivalent AHS:

but this works:

2025-05-04 update: I didn't bring enough attention to the AHS duals because they are the key to this. Recently YZF posted this puzzle to the forums:

It contains an AHS-XZ with no equivalent ALS-XZ. 

(169)(r9c8 = r134c8) - (2349)(b6p2 = b6p1367) => r8c7<>9

Let's look at the intersection r4c8 more closely. The candidates are (1246). (16) is part of the red AHS, (24) is part of the green AHS. The cell link connects these two AHS and prevents them both from occupying this cell, hence at least one of them must not contain this cell and must contain every other cell in the AHS. What about as an ALS-XZ?

This looks familiar...

(249)r479c8 = (12-46)r4c8 = (169)b6p279 => r8c7<>9

Yes it's exactly the same case as before. This makes it clear that the "Dual AALS link" is not worth considering if you are already including AHS in your chains, because it has an AHS counterpart that doesn't require any funky grouped links