Yes, I knew the cardinalities in question were finite. The point applies regardless though. For any set X, there is no injection from 2^X to X. In the finite case, this is 2^n > n for all natural numbers n.
If there are N possible states, then the number of functions from possible states to {0,1} is 2^N , which is more than N, so there is some function from the set of possible states to {0,1} which is not implemented by any state.
Yes, I knew the cardinalities in question were finite. The point applies regardless though. For any set X, there is no injection from 2^X to X. In the finite case, this is 2^n > n for all natural numbers n.
If there are N possible states, then the number of functions from possible states to {0,1} is 2^N , which is more than N, so there is some function from the set of possible states to {0,1} which is not implemented by any state.
I never said it had to be implemented by a state. That is not the claim: the claim is merely that such a function exists.