Dylan Thurston writes Can you say a bit more about the proof? I would guess that essentially from the definition of tensor product you can get an algorithm that possibly uses negative area rectangles, and the difficulties are avoiding those? This is exactly correct. Our secret weapon is, POSITIVE BASIS THEOREM. Let {a_1, . ,a_n b} be a set of positive (real) numbers. Then there exists a set {b_1,. . ,b_m} of positive numbers such that 1. The b_i are rationally independent, 2. The a_i are rational linear combinations of the b_i with NON-NEGATIVE coefficients. Such a simple statement! Has anyone seen it before? I won't spoil your fun by giving a proof. I'll try to explain referring to the example below. Verify the rectangles above and below the line have the same symmetric (that is a(x)b = (b(x)a) tensor area * * * * * * * * * * * * * * * * * b-d * * d * * * * * * * * * * * * * * * * * * * a a-c ____________________________________________________________ * * * * * * * b-c * * * * * * * b * * c * * * * * * * * * * * * * a-e * * a-e-c * * * * e In the Example the heights and widths of the 5 rectangles are the {a_i} and a basis might be {a-c-e, b-c,c-d, d,e}. you can verify condition 2. Next in 2. replace let b'_i = b_i/d where d is a common denominator of the coefficients. Now all widths and heights are non-negative INTEGRAL combinations of the b'_i. OK, for each rectangle R, sides a and b are integer combinations of the b'_i so R can be tiled as a cartisian product with tiles of form b'_i(x)b'_j but these form a basis for the tensor product, so if two families of rectangles have the same tensor area then the expression for this tensor area in terms if the basic ij tiles is the same, and hence for all i and j the two tilings have the same number of ij tiles for each ij. done