. Proof, ?. Let, . R. ?-s-be-incorporable-w, . ???, ?. ?. Let et al., By ?'s incorporability, there is a context ? 0 ? ? (perhaps not in ?) such that ? ? 0 = ? ? ?{?? ?} and % ? 0 is faithful to % ? . By independence between outcome and state awareness

X. ?. As and S. ?. , the last identity holds because ? ? 0 refines ? ? ), we have ? ? ?. So it remains to show two things

?. ?. , ?. ?. , and ?. Is, To complete the proof of Ax. 6 for the substructure, it suffices to note that (i) ? 0 ? ? because S ? 0 = S ? (as ? ? 0 = ? ? ? {? 1 ? ???? ? ? }), and (ii) S ? is partitioned into (the non-empty sets among) ? 1 ?, Proof of Lem. 32. Assume Ax. 1?6. Let R, (? ? ) ??? , and (? ? ) ??? be as specified. Each ? ? induces a function ? ? on R via ? ? (?) := ? ? (? ? S ? ) (? ? R)

?. ?. , ?. ?. Each, ?. ?-is-in-r|-s-?-;-so-we-write, and ?. ?. , Second, I show fineness. Let ? ? 0. As ? ? is fine, we may partition S ? into ? 1 ? We may take ? 1 ? ???? ? ? to partition S, by the argument in fn ) ? ? for all ? ? . So ? ? is fine. Claim 2 : ? ? is the same for all ? ? ?. Let ?? ? 0 ? ?; we show that ? ? = ? ? 0 . By Claim 1 and Lem. 16 and 17, it suffices to show that ? ? and ? ? 0 are ordinally equivalent. Let ?? ? ? R. As ? and ? are incorporable, we may pick a context ? ? ? in which ? and ? are representable, The events ? ? ? ? ? (? ? ? ) representing ? and ? respectively also represent ? ? S ? and ? ? S ? respectively. Now (*) ? ? (?) ? ? ? (?) ? ? ? % ? ? ? ? since ? ? (?) ? ? ? (?) ? ? ? (? ? S ? ) ?

?. ?. , ?. , ?. ?. , ?. Represent, ?. ?. et al., Analogously, as ? and ? are incorporable we may pick a context ? 0 ? ? 0 where ? and ? are representable; as before), as required. Claim 3 : The (by Claim 2 ?-independent) probability measure ? :? ? ? is agreed among the ? ? ( ? ? ? ) For any ? ? ? , recall that ? * ? is the function of (representable) objective events ? ? S induced by ? ?, **) ? ? 0 (?) ? ? ? 0 (?) let ? * * Condition (b) holds because ? ? ? {?? ?} = ? ? ? {?? S ? \?}. Condition (c) holds because, as ? ? S ? , we have ? * * ? (?) = ? * ? (?) and ?(?) = ? ? (?)

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