Ok, so how about this...

b is the same as (!g ^ !s ^ !b) V (g ^ !s ^ !b) V (!g ^ s ^ !b) V (!g ^ !s ^ b)

So if b is true then so is (!g ^ !s ^ !b) V (g ^ !s ^ !b) V (!g ^ s ^ !b) V (!g ^ !s ^ b)

So I can use

AND here and say b AND [(!g ^ !s ^ !b) V (g ^ !s ^ !b) V (!g ^ s ^ !b) V (!g ^ !s ^ b)]

And I can throw out any !b, since I can't have b and !b at the same time.

So it simplifies to (!g ^ !s ^ b)

Which, using "Distributivity over ^", would give me (!g ^ s) ^ (!g ^ b), So, the apple is both in the silver container and the bronze container, that cant be right...