Exhibit 3

Proof That X²M Avoids Row Preference
Reversals (RPRs) for Strong Form Row Dominance

I. Definitions

(1)	X2M Model:	Minimize X2 = åi åj ( ( rij - yi / yj ) 2 / ( yi / yj ) ) = åi åj ( rij - yij ) 2 yji
		Subject to:
		1 = åiyi 0 < yi for i = 1,...,n
(2)	X2ij = ( rij - yij ) 2 yji + ( rji - yji ) 2 yij + åk ¹ i, j ( ( rik - yik ) 2 yki + ( rjk - yjk ) 2 ykj + ( rki - yki ) 2 yik + ( rkj - ykj ) 2 yjk )
(3)	X2 = X2ij + X-2ij º ( X2 components having a subscript i or j ) + ( All other X2 error components ) for solution vector [ y ] = [ y1,...,yn ]
(4)	X2i/j = ( rij - yji ) 2 yij + ( rji - yij ) 2 yji + åk ¹ i, j ( ( rik - yjk ) 2 ykj + ( rjk - yik ) 2 yki + ( rki - ykj ) 2 ykj + ( rkj - yki ) 2 yik
(5)	X2/ = X2i/j + X-2ij º ( X2 components having a subscript i or j ) + ( All other X2 error components ) when yi and yj are transposed in [ y ]
(6)	DX2ij = X2 - X2/ = X2ij = X2i/j º Change in X2 resulting from transposing yi and yj in [ y ] .
(7)	[ y * ] = [ y1 *,...,yn * ] º Optimal X2M solution vector
(8)	yij = yi / yj º X2M model surrogate for the rij response judgment

II. Theorem

If r_ik > r_jk for k = 1,...,n

Then y_i * > y_j *

Interpretation: If Row i is dominant in the strong form over Row j, then the optimal X²M solution will not reverse the Row i dominance over Row j.

III. Proof

DX2ij	=	X2ij - X2i/j
	=	rij2 ( yji - yij ) + rji2 ( yij - yji ) + åk ¹ i, j ( rik2 ( yki - ykj ) + rjk2 ( ykj - yki ) + rki2 ( yik - yjk ) + rkj2 ( yjk - yik ) )
	=	( rij2 - rj2i ) ( yji - yij ) + åk ¹ i, j ( ( rik2 - rjk2 ) ( yki - ykj ) + ( rkj2 - rki2 ) ( yjk - yik ) )
If yi < yj when rik > rjk for k = 1,...n
Then ( rij2 - rjk2 ) ( yji - yij ) > 0, ( rik2 - rjk2 ) ( ykj - yki ) > 0, and ( rkj2 - rki2 ) ( yjk - yik ) > 0
\ DX2ij = 0 if rik = rjk for all k = 1,...,n
\ DX2ij = 0 if rik > rjk for at least one k among k = 1,...,n

Conclusion: Conditions r_ij > r_jk for all k, r_ik > r_jk for at least one k, and y_i < y_j Þ y_i ¹ y_i * and/or y_j ¹ y_j *. Q.E.D.

IV. Interpretation

In any response matrix [ R ] whose r_ik elements have a reciprocal property and r_ik > r_jk for all k = 1,...,n, then the optimal X²M solution vector will always have components in which y_i * > y_j *. In other words, if Row i strongly dominates Row j, then any solution in which y_i < y_j cannot be X²M optimal since X² can be reduced simply transposing y_i and y_j in the solution vector.

Whenever Row i and Row j have identical r_ik = r_jk values for all k = 1,...,n, then no error change can be achieved since DX²_ij = 0. However, in this case y_i * = y_j * in a X²M optimal solution.