Shortest-proof-game challenge

That is indeed a problem. Since Bxh7 was a condition that had to be met, I hadn't thought about the bishop moving in the other direction. Let me fix that in a few moments.

As it turns out, I have to resort to a completely different strategy. This time there should be no holes. Can anyone find a different method?

It's a fairy proof game, not a mating puzzle.

This feels a fun position to play around with.

I believe this is legal.

That looks right to me. My comment about "easily be changed" was a little off base.

does not belong here

Actually, the reflexmate solution does not work completely as there is a contradiction between the two challenges. You can't legally play Ke2 when there is a checkmate alternative available. Also the rook on f1 need not give check to force the black checkmate; the reflexmate requirement takes care of it. That probably allows more and different proof game constructions.

The only formal way to save it is by simply declaring that the reflexmate rule ends at the challenge fork. In that sense this challenge is precisely the one you want to avoid in a reflexmate setting

Or one could give the requirement for the end position, as in the original challenge, and simply add the condition that it be reached without White having a previous opportunity for mate in one, without using the term reflexmate.

Then Leither could give a solution and there would be no need to decide whether it could be called a reflexmate solution. As Arisktotle said, the challenge required a position where a move could occur instead of an available mate in one.

@5733
I believe this is illegal.
The last move must have been ...Bf7xNg8.
The white king cannot have entered from the right because of the black pawns, so he must have taken a route from the left, say Ke1-d2-c3-c4-c5-c6-d7-e8xBf8-g8-h8. That is one capture of Bf8.
For the white pawn to land on e5 it must have captured twice: c2-c3xd4xe5 that is 2 captures.
That makes a total of 1 + 2 = 3 captures of black men.
There are still 14 black men on the board.
14 black men on the board + 3 captured = 17 men > 16 men
Hence illegal.

#5732 is illegal. The black king could never have gotten to h2 with the position of the white pawns. The same way a king can never get to the back rank with 4 pawns still on the 7th rank like this:

Ke8-d8-c8-b7-b6-a5-b4-c3-d2-e1-f1-g1-h2... looks good to me.

Thanks nice work!!!

Illegal, but your reasoning is flawed. 14 Black pieces are present and 2 are missing, enough for the White pawn capture. However, bBf8, due to the f7 and g7 pawns being home, never could have moved to let it out to be captured

Proof game: quickest 'Bdxf6+'

Your game is shorter than Kyobir's game, but if it's required that you can't also say it another way, it might require another move (in this position, you could also say B4xf6.)

Guys I am not sure of the position I posted earlier, is legal:

I tried to construct a game for this thread to get this position and the closest I got was needing 2 more captures to get 1 of the pawns to the d or e files. There may not be enough captures. Am I correct?

it's that one puzzle where you have to go h3, h4, h5... h8=N... there is a smothered mate somewhere

@5746

This has been proven legal before. See @847