Here in 09-20's puzzle the pentagonal situation diagram is :
 in which

thick lines indicate that one exactly of the vertices is true,

thin lines indicate that one at most of the vertices is true.

[Maple Plot]

Check this stands in our puzzle: (i8=2) and (i3=2) can't be both true (only a 2 in Ci), 2 is exactly in one of the two cells i3, e3 (no other place for 2 in R3), (e3=2) and (e9=2) can't be both true (only a 2 in Ce), 2 is exactly in one of the two cells e9, d8 (no other place for 2 in Be8), (d8=2) and (i8=2) can't be both true (only a 2 in R8). How many words shortened in a single diagram...

[Maple Plot]

And it's a general result : when a pentagonal situation arises between assumptions ABCDE, the vertex A is always false.

You can argue that by " If A were true, then B and E would be false, so C and D would be true : impossible ", or " whether B or C (in which D is false, E is true), A is false ", or anything else, the result is clear : when we have pentagonal ABCDE, A is false.

Note that "Xwing" and "Two values in two cells in same unit" are special cases of pentagonal situations.
Note also that pentagonal situations generalize in hepta, nona, ... odd-agonal situations with the same result : the only vertex from which leave two thin lines is always false.

Close this window and you're back to 09-20's proof.