Illegal Position Contest!

I have got a proof that cobra91's position from #171 is illegal:

Because white has two rooks and there is no way for white's a-rook to have gotten out of the pawn chain, it is clear that white's a-rook has been captured and white's h-pawn has promoted to a rook. Because black has still got his d,e,f,g,h pawns, the white h-pawn has moved to the c7-square and has therefore promoted either on b8,c8 or d8. As black's LSB has never left the c8 square, the h-pawn didn't promote on c8 and also didn't promote on d8 (because then the rook would have no chance to get to a8 or b8). Therefore the white h-pawn has promoted to a rook on the b8 square. On its way there, it has captured 6 black pieces/pawns.

It is obvious that the h-pawn didn't capture the black h-rook, the black king, black bishops and the b,d,e,f,g,h pawns. The only remaining pieces/pawns that the h-pawn might have captured are the a,c pawn, the black queen, a-rook, and 2 knights. These are 6 black pieces/pawns in total and thereby the white h-pawn must have captured all of these 6 pieces/pawns.

Because the white h-pawn and the black a-pawn cannot cross paths (without promoting), the black a-pawn must have promoted. In order to do so, the black a-pawn must have captured a white piece. As white has still 15 pieces remaining, it is clear that the black c-pawn did not capture any white pieces. Hence, the black c-pawn did not promote.

Then, however, the only possible way that the white h-pawn captured the black c-pawn on its march to b8 is if the black c-pawn didn't move away from its starting position on c7.

Then, before the white h-pawn captured the c-pawn on c7, there is no way for the black queen to leave the fields d8-h8. This especially means that on its march to c7, the white h-pawn didn't capture the black queen and as such, the black queen is the last black piece/pawn that was captured by the white h-pawn in order to promote.

But then, however, it is obvious that the white h-pawn captured the black queen on b8. But because the black c-pawn has never moved (and neither did the c8-bishop), it is obvious that the black queen cannot have moved to the b8 square. Contradiction. Hence, the position is indeed illegal.

(please excuse my bad English)

Well done! This logic is bulletproof. 

Pilchuck was close, but one cannot conclusively prove illegality without addressing all potential pawn promotions (see post #180), as was done in the above analysis.

Um, Cobra, I did address all the possible pawn promotions, you just ignored it.

^ I did no such thing!   I replied to your most recent comment in post #180; here's the quote:

The significance is that White can, in fact, get his pawn all the way to c7, and still seemingly have another Black piece to capture (in order to promote). So the issue is more subtle than merely a lack of available captures.

Since the details of my earlier position have now been worked out (Bravo, yellowchesstiger! ), I'll submit another which is interesting... and lies just inches outside the fringes of legality.

#199 might be an interesting starting position for a game.  The computer evaluates it as pretty even.

it is illegal because of the doubled pawns

I have got a proof that cobra91's position from #197 is illegal:

Assume that the desired position P was legal with player X to move.

We first proof the following lemma:

Lemma 1: From the last three (half-)moves that led to position P, no one of them was Nxf7.

Proof: If Lemma 1 was false, then there is no way for white to have a white LSB on e8 in Position P. qed

Now we proceed to proof:

Lemma 2: X=White (meaning that in Position P it is White to move)

Proof:

If X=Black (meaning it is Black to move in position P), we ask ourselves what white's last move was (white's last move does exist because the position is not the starting position). The only white piece/pawn that could performed white's last move is the Nf7, as all the other white pieces and the Ph4 cannot have moved to the square on which they are sitting in the position P, because all potential squares from which they could have reached their squares are blocked, respectively (and white's other pawns haven't moved at all during the game). But if the Nf7 performed white's last move, the black king would have been in check (by the Be8) during white's turn (because Nxf7 is impossible according to Lemma 1), contradicition .

Hence, in the Position P it is white to move (meaning that X=White). qed

Now we proceed to proof the following Lemma:

Lemma 3: Black's last move that led to position P was either fxg5 or Nxg3.

Proof of Lemma 1:

0)  Assume Lemma 3 is false.

1) According to Lemma 2, in the Position P it is white to move (meaning that X=White).

2) We now seek to determine what black's last move m was (the move leading to position P). Let P_1 be the position in which black moves m to reach position P.

If the move m was not performed by the Ng3 or the Pg5 (in position P), then we again obtain a contradiction analogously to 1) if we want to find out what white's move was which led to position P_1.

Hence, we have to discuss the following four cases:

case 1: Black's last move m was g7-g5.

case 2: Black's last move m was g6-g5.

case 3: Black's last move m was Ng3.

case 4: Black's last move m was a capture (the only remaining options are fxg5 or Nxg3).

Now, one immediatly realises that case 1 is not possible for obvious reasons (because then there is no legal white move leading to P_1 (except for Nxf7, but this contradicts Lemma 1.); this can be seen with analogous arguments like in section 1).)

One also notices that case 3 is not possible: If case 3 was possible, because Ng3 is not a capture move, the white move n that lead to P_1 was performed by one of the white pieces that is still on the board in position P. For identical reasons listed as in section 1), however, we realise that the only white piece that could have legally performed move n is the Nh1 in position P. Hence, the move n looked like this: Nh1-g3. But then Ng3 is a capture move, contradiction.

We also notice that case 2 is impossible: If case 2 was the case, we notice that the white move n which led to position P_1 must have been performed by the Nf7 as all other white pieces are "blocked" (meaning there exists no square from which they could have moved to the square which they sit on in Position P_1). But if move n was performed by the Nf7, the move g5-g6 would leave black in check, if Nf7 wasn't a capture: Nxf7. But Nxf7 again contradicts Lemma 1.

And last but not least, case 4) obviously contradicts assumption 0).

This ends the proof of Lemma 1. qed

The following Lemma 4 is more or less obvious:

Lemma 4: a) White and Black have both promoted a pawn to a light-colored bishop.

b) Neither White's nor Black's LSB in the initial starting position was captured by an opponent's pawn on its march to promotion.

Proof:  In the starting position, neither the white nor the black LSB could have left their initial starting position. qed

In the following section we will now seek to construct a contradiction to the piece count left on the board.

Lemma 5: White's c-pawn promoted to a LSB. White's a- and b- pawn never left their file.

Proof: According to Lemma 4, at least one of the white a,b,c pawns promoted to a LSB.

If the white a or b pawns promoted to a LSB, they both would have needed to capture at least fie black pieces/pawns in order to reach one of the two white promotion fields (e8, g8) from which a bishop can go to e8, eventually. Because black still has got 11 pieces left and the Bc8 in the initial starting position wasn't captured by the white pawn that promoted according to Lemma 4 b), black must have had at least 11+5+1>16 pieces when the game started, contradicition.

Hence, White's c-pawn promoted (and White's c-pawn is the only white pawn that promoted). In order to reach one of the two possible promotion squares e8 and g8, the white c-pawn is required to capture 4 black pieces/pawns. Because 11 black figures are still left in position P and because the black LSB on c8 wasn't captured by a white pawn and because 11+4+1=16 and black only has 16 figures to start with, we obtain that neither White's a nor White's b pawn captured a black figure and that White's c-pawn captured exactly 4 black figures on its march to promotion.  qed

Because according to Lemma 5 the white c-pawn captured exactly 4 black figures, it didn't capture Black's h-pawn on the h-file.

Hence, because in P there is no black h-pawn, it is clear that either Black's h-pawn was captured by a white figure on  the h-file or that the Black h-pawn captured a white piece in order to leave the h-file. If, however, the black h-pawn was captured on the h-file, we can count some the black pieces of the initial position in the following way: 1x h-pawn, 4x figures captured by white c-pawn, 11xfigures still left in position P, 1xBlack LSB on c8. This adds up to 17 black pieces, contradiction. Hence, we have obtained the following lemma:

Lemma 6: The black h-pawn captured a white figure.

Proof: see above. qed

Now, we will proceed to prove:

Lemma 7: The black pawn that promoted captured 4 or 5 white figures, but not white's a-pawn. White's a-pawn was captured by another black figure.

Proof: One of black's f,g,h pawns must have promoted on a light square to a LSB. As a bishop cannot reach f3 from the fields h1 and f1 (due to being blocked by white pawns), it is clear that a black pawns must have promoted to a LSB either on b1 or on d1. Either way, it must have captured at least 4 white pieces/pawns on its march to promotion.

Because the White a-pawn never left the a-file (Lemma 5), if the Black pawn that promoted would have captured the a-pawn, it would have had to capture another white figure to get back onto the b-file to become a LSB. Hence, it would have captured at least 6 white figures. In P, there are 9 white figures left. If one also considers the White LSB which was on f1 and the white figure that was captured by black one move before position P was reached (Lemma 3), we obtain 9+6+1+1=17>16 White figures, contradicition.

Hence, the black pawn that promoted didn't capture the white a-pawn, and therefore captured 4 or 5 white figures different from the a-pawn.

Because White has no a-pawn in position P, the white a-pawn must have been captured by another black figure. This ends the proof of Lemma 7. qed

Lastly, we prove:

Lemma 8: The black h-pawn didn't promote to a LSB.

Proof: If it did promote to a LSB on squares b1 or d1, it would have needed 6 captures to get there. This contradicts Lemma 7. qed

Now, we have anything we need in order to prove that position P is indeed illegal:

- in P, there are 9 white figures

- on the black move leading to position P, a white figure was captured (Lemma 3)

- the black h-pawn did at some point of time capture a white figure (Lemma 6)

- the black pawn that promoted captured at least 4 white figures (Lemma 7), these have not been counted yet (Lemma 7,8)

- the white a-pawn was captured (Lemma 5)

- the white LSB on f1 was captured (Lemma 4)

This, however, adds up to 17 white figures (we have not counted anything twice). Contradiciton. This ends the proof.

…I think.

The position in post #209 is legal: