> Each piece and pawn can maximally make 6 half-moves
This is NOT correct, however, the idea seems right to me.
It may be possible that, say, b2 pawn is captured, so b1 knight may do 7 half-moves, not 6.
Let us consider last position. It must have holes (it may not be 8+8+8+8 pieces), and one have to ask question "where such pieces disappeared". This is the key question.
To #11, I suggested that captured are only pawns and pieces of one side, so second side can move ALL its 16 pawns and pieces to 7 and 8 rows. If for example b2 pawn is captured, than White has only 15 pawns and pieces which gives them only 15*6 = 90 (maybe +1) half-moves which is already not enough.
You forgot that b2 pawn may be captured not on b2, but on b6. That is we have 14*6 + 4 + 7 = 95 moves, not 91
4 moves are b2-b3-b4-b5 and 7 moves are Nb2...Nb8 (really knight may not go like this, but it may be swapped with another piece)
The rest is to prove that there is not possible b2-b3-b4-b5-b6, that still give 96 moves.
Sorry, correct version is:
4 moves are b2-b3-b4-b5-b6 and 7 moves are Nb2...Nb8 (really knight may not go like this, but it may be swapped with another piece)
The rest is to prove that there is not possible b2-b3-b4-b5-b6-b7, that still give 96 moves.
I suspect that 96-4=92 moves is really possible.
and 7 moves are Nb1...Nb8, of course... hope you grasp an idea
If b2 pawn is captured on b6 or even on b7 (which is reasonable), than black pawn on b7 must be captured on its place, b7 - which gives lack of half-moves already to black (theoretically 15*6+1).
I think, the point is impossibility to keep ALL 32 pawns and pieces of both sides to make needed 6 * 32 = 192 half-moves.
You again are not correct, try to understand why.