This is part of of D52508. It uses the demand bits of umax(A, C) to prove we use just A, so long as:

"The lowest non-zero bit of DemandMask is higher than the highest non-zero bit of C"

https://rise4fun.com/Alive/q4Z

Conceptually, this works because:

if A > C, we pick A. No big deal.

if A < C then all the demanded bits of A == demanded bits of C, so we can pick either and get the same result. So we pick A.