I am trying to check that I am understanding this correctly by applying it, though probably not in a very meaningful way:
Am I right in reasoning that, for S⊆W , that 1S⊲C iff ( (C can ensure S), and (every element of S is a result of a combination of a possible configuration of the environment of C with a possible configuration of the agent for C, such that the agent configuration is one that ensures S regardless of the environment configuration)) ?
So, if S = {a,b,c,d} , then C=⎡⎢⎣abacddefa⎤⎥⎦
would have 1S⊲C , but, say
D=⎡⎢
⎢
⎢
⎢⎣aaabbccdfbfa⎤⎥
⎥
⎥
⎥⎦
would have 1S⋪D , because , while S can be ensured, there isn’t, for every outcome in S, an option which ensures S and which is compatible with that outcome ?
There is a single morphism from 1S to ⊥ for every world in S, so 1S◃C means all of these morphism factor through C.
A morphism from C to ⊥ is basically a column of C and a morphism from 1S to C is basically an row in C, all of whose entries are in S, and these compose to the morphism corresponding to the entry where this column meets this row.
Thus 1S◃C if and only if when you delete all rows not entirely in S, the resulting matrix has image S.
I think this equivalent to what you said. I just wrote it out myself because that was the easiest way for me to verify what you said.
I am trying to check that I am understanding this correctly by applying it, though probably not in a very meaningful way:
Am I right in reasoning that, for S⊆W , that 1S⊲C iff ( (C can ensure S), and (every element of S is a result of a combination of a possible configuration of the environment of C with a possible configuration of the agent for C, such that the agent configuration is one that ensures S regardless of the environment configuration)) ?
So, if S = {a,b,c,d} , then
C=⎡⎢⎣abacddefa⎤⎥⎦
would have 1S⊲C , but, say
D=⎡⎢ ⎢ ⎢ ⎢⎣aaabbccdfbfa⎤⎥ ⎥ ⎥ ⎥⎦
would have 1S⋪D , because , while S can be ensured, there isn’t, for every outcome in S, an option which ensures S and which is compatible with that outcome ?
Yep.
There is a single morphism from 1S to ⊥ for every world in S, so 1S◃C means all of these morphism factor through C.
A morphism from C to ⊥ is basically a column of C and a morphism from 1S to C is basically an row in C, all of whose entries are in S, and these compose to the morphism corresponding to the entry where this column meets this row.
Thus 1S◃C if and only if when you delete all rows not entirely in S, the resulting matrix has image S.
I think this equivalent to what you said. I just wrote it out myself because that was the easiest way for me to verify what you said.
Thanks! (The way you phrased the conclusion is also much clearer/cleaner than how I phrased it)