Suppose one has three polygons, A, B, and C, which lie along the same projector from the screen. Suppose A and B are transparent and one models this with interpolated transparency:

We will show that for this model, one gets different pixel values for the image of the opaque polygon C depending upon whether A is in front of B or vice versa.

Given: Polygons A, B, C; polygon opacity factors: K_A, K_B, K_C; and polygon intensities (perhaps RGB triplets: I_A, I_B, I_C)

A, B, and C lie along the same projector with C being the last. C is completely opaque so its opacity factor is K_C = 1. Now using the interpolated transparency equation, show that the order A, B, and then C results in a different intensity than B, A, and then C. So, let's solve for the order A, B, C and then interchange A and B to show inequality (rI_K = resulting intensity at polygon K):

rI_A = (1-K_A)I_A + K_ArI_B
rI_B = (1-K_B)I_B + K_BrI_C
rI_A = (1-K_A)I_A + K_A[(1-K_B)I_B + K_BI_C]

now swapping the intensity and opacity factors for polygons A and B:

rI_B = (1-K_B)I_B + K_B[(1-K_A)I_A + K_AI_C

If K_A!=K_B and I_A!=I_B, we can prove that iRA!=iRB (resultant intensity with polygon A in front does not equal resultant intensity with polygon B in front).

Our goal is to show that the resultant intensities are not equal for a different ordering of polygons
(rI_A!=rI_B). So if rI_A=rI_B , they must be equal for all values of K_A, K_B , I_A, and I_B; therefore finding just one example that shows otherwise proves their inequality. Here goes...let's try:

I_A = 1, I_B = 0.5, K_A = 0.2, K_B = 0.5 : (I_C = 1)
rI_A = (1 - 0.2)(1) + (0.2)[(1 - 0.5)(0.5) + (0.5)(1)] = 0.95
rI_B = (1 - 0.5)(0.5) + (0.5)[(1 - 0.2)(1) + (0.2)(1)] = 0.75

CLEARLY NOT EQUAL, therefore rI_A!=rI_B by counterexample.