ronk wrote:In the world of cover set logic if, for any exocet pattern digit, there exists 1) a column void of candidates (as in the 2nd diagram above), or 2) less than two rows with less than two candidates (in the three columns as in the 3rd diagram above), those candidates becomes irrelevent to the logic set. Either 1) or 2) occurs in 41 of the 42 puzzles listed here.

A much simpler approach is possible:

For a candidate (a) that's true in the base cells, the JExocet pattern limits it to just two target cells in intersections of the 3 cross-lines with the JExcoet band.

Condition 3 limits the 18 external cross line cells to holding a maximum of two instances of (a), so ensuring it must occupy at least one of these target cells.

With two true base digits subject to this limitation, each one must occupy a different target cell, so eliminating all non-members from both of them.

There is therefore no need to resort to different SLG diagrams each time to prove this.

ronk wrote:I know what you're trying to say, but I find those diagrams confusing. If you're going to show only six in the left, why show more than four in the others?

I was showing the most cells that could be occupied in the cross lines in each case.

ronk wrote:Does that mean you don't consider base r9c13 and target r8c89 to be an "Exocet" either?

That doesn't comply with the JExocet pattern requirements where the base cell pair and the two target cells must all be in different boxes in the same band.

By champagne's rules, if combining the template options for a set of 3 or 4 digits succeeds in showing that two cells must hold the same digits as a selected pair of cells, that arrangement is an Exocet pattern, otherwise it's not. For an Exocet these cells are identified as targets where non-member digits can be eliminated. I therefore feel unqualified to answer.