SpAce wrote:(2=79)r2c2-(79=4)r2c39-(4=1)r4c3-(1=2)r4c2

That does make sense if the first instance of "79" is read as "7 or 9" and the second as "7 and 9". For some reason that way of reading it wasn't intuitive to me.

To avoid misunderstandings, we write -(7|9=2)r2c2. If 7 or 9 is missing in 479/479, the 4 must be there ...

I would write the loop as (2=4)r4c23-(4=2)r2c239, loop.

This is also a double-linked ALS-XZ.

Another approach is:

You have 5 digits in the 5 cells. None can be in 2 of the cells. So eliminate all in the rest of their units.

In JC's second example you have 5 digits 14678 in 6 cells.

Only the 7 can be in 2 of the cells. Eliminate 1468 from the rest of their units.

Then look, where the 7's can go to, and you will find more eliminations.