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. Proof, It follows from Propositions 5

. Proof and . Clearly, We are going to show that it is faithful M od(?) then s k (M od(?), )) = (1, . . . , 1) by definition of s k . So we cannot have (M, w) < ? (M ? , w ? )

?. So and ?. , jn such that j 1 = j 0 , j i = j i+1 , ?(R j 0 ? R j 1 ?, ? R jn (w ?? ), R j 0 ? R j 1 ? . . . ? R jn (w)) = 1 ? ?(R j 0 ? R j 1 ? . . . ? R jn (w ? ), R j 0 ? R j 1 ? . . . ? R jn (w))

?. N. Besides and ?. , jn such that j 1 = j 0 , j 1 = Y , j i = j i+1 , ?(R j 1 ?, ? R jn (w ?? ), R j 1 ? . . . ? R jn (w)) = ?(R j 1 ? . . . ? R jn (w ? ), R j 1 ? . . . ? R jn (w)) because M ??

@. .. @bullet-r-jn, @. @bullet, ?. G. , and =. , ? R jn (w)) | j i = j i+1 , j 1 = Y } ? m{? ? R jn (w)) | j i = j i+1

S. ?? and ?. Od, Then (M ? , w ? ) / ? M od(? ? ?) which is impossible by assumption