Proof of Proposition 5 Sample Clauses

Proof of Proposition 5. Choose Kj to be a subgroup of K minimal with the property that KjΩ1 ∼= Alt(d). Then

Proof of Proposition 5. Since the Gaussian model satisfies the condition in Theo- rem 5, it suffices to evaluate C = I(x, s; yr|ye). I(x, s; yr|ye) = h(yr|ye)− h(yr|ye, x, s) (46) = h(yr|ye)− h(zr|ze) (47) = h(u|ye)− h(u|yr) 1 . √ (P + Q + 2√PQ)2 Σ 1 Further evaluating each of the terms above with u = x + αs, note that = 2 log 2πe P + Q + 1 + 2 PQ − P + Q + 1 + 2√PQ + ∆

Proof of Proposition 5. Like for strong equivalence in Proposition 8.2.1 from [46], we shall prove the following fact about step stochastic bisimulation. Let us have ∀j ∈ J , Rj : G↔ssG′ for some index set J . Then the transitive closure of the union of all relations R = (∪j∈J Rj)+ is also an equivalence and R : G↔ssG′. Since ∀j ∈ J , Rj is an equivalence, by definition of R, we get that R is also an equivalence. Let j ∈ J , then, by definition of R, (s1, s2) ∈ Rj implies (s1, s2) ∈ R. Hence, ∀Hjk ∈ (DR(G) ∪ DR(G′))/Rj , ∃H ∈ (DR(G) ∪ DR(G′))/R, Hjk ⊆ H. Moreover, ∃J ′, H = ∪k∈J ′ Hjk. We denote R(n) = (∪j∈J Rj)n. Let (s1, s2) ∈ R, then, by definition of R, ∃n > 0, (s1, s2) ∈ R(n). We shall prove that R : G↔ssG′ by induction on n. It is clear that ∀j ∈ J , Rj : G↔ssG′ implies ∀j ∈ J , ([G]≈, [G′]≈) ∈ Rj and we have ([G]≈, [G′]≈) ∈ R by definition of R. It remains to prove that (s1, s2) ∈ R implies ∀H ∈ (DR(G) ∪ DR(G′))/R, ∀A ∈ NLfin, PMA(s1, H) = PMA(s2, H). • n = 1 In this case, (s1, s2) ∈ R implies ∃j ∈ J , (s1, s2) ∈ Rj. Since Rj : G↔ssG′, we get ∀H ∈ (DR(G) ∪ DR(G′))/R, ∀A ∈ NLfin, PMA(s1, H) = ∑ PMA(s1, Hjk) = ∑ PMA(s2, Hjk) = PMA(s2, H). • n → n + 1 k∈J ′ k∈J ′ Suppose that ∀m ≤ n, (s1, s2) ∈ R(m) implies ∀H ∈ (DR(G)∪DR(G′))/R, ∀A ∈ NfiLn, PMA(s1, H) = PMA(s2, H). Then (s1, s2) ∈ R(n + 1) implies ∃j ∈ J , (s1, s2) ∈ Rj ◦ R(n), i.e. ∃s3 ∈ (DR(G) ∪ DR(G′)), such that (s1, s3) ∈ Rj and (s3, s2) ∈ R(n). Then, like for the case n = 1, we get PMA(s1, H) = PMA(s3, H). By the induction hypothesis, we get PMA(s3, H) = PMA(s2, H). Thus, ∀H ∈ (DR(G)∪DR(G′))/R, ∀A ∈ NLfin, PMA(s1, H) = PMA(s3, H) = PMA(s2, H). R ↔ R R By definition, ss(G, G′) is at least as large as the largest step stochastic bisimu- lation between G and G′. It follows from above that ss(G, G′) is an equivalence and ss(G, G′) : G ssG′, hence, it is the largest step stochastic bisimulation between G and G′.

Proof of Proposition 5. 4.1. First suppose that each of the subquo- tients Mi/Mi+1 are of rank one. If M is of rank one itself then we have already seen that M admits a unique crystalline GK-action. For general M we have an exact sequence Arguing inductively on the length of M we may assume M1 and M/M1 admit unique crystalline GK-actions such that each Mb ⊂ Mb is GK-stable.