Counterexample to your claim that Ctrl(C⊕D)=Ctrl(C)∪Ctrl(D) (with both C and D not null):
Take C and D to be the following Cartesian frames over world {x,y} with agent and environment both {a,b} as the following matrices:
Then Ctrl(C)=Ctrl(D)=∅ but Ctrl(C⊕D)∋{x}.
(Sorry for my formatting—I haven’t done LaTeX in these comments before.)
Thank You! You are correct. Oops!
I deleted that claim from the post. That was quite a mistake. Luckily it seems self-contained. This is the danger of “Trivial” proofs.
Initial versions of the theory only talked about ensurables, and the addition of (and shift of attention to) controllables came much later.
Counterexample to your claim that Ctrl(C⊕D)=Ctrl(C)∪Ctrl(D) (with both C and D not null):
Take C and D to be the following Cartesian frames over world {x,y} with agent and environment both {a,b} as the following matrices:
abayxbxxabaxybyyThen Ctrl(C)=Ctrl(D)=∅ but Ctrl(C⊕D)∋{x}.
(Sorry for my formatting—I haven’t done LaTeX in these comments before.)
Thank You! You are correct. Oops!
I deleted that claim from the post. That was quite a mistake. Luckily it seems self-contained. This is the danger of “Trivial” proofs.
Initial versions of the theory only talked about ensurables, and the addition of (and shift of attention to) controllables came much later.