Lemma. The above protocol achieves King Consistency and Validity remains. Proof.
Lemma. Theorem 3 [k-Set Agreement] The protocol solves the k-set agreement problem. Proof The proof follows from the Lemmas 1, 2, and 3. T heorem Theorem 4 [Early Stopping] No process halts after the round min bf kc bt kc . Proof Let us first observe that a process decides and halts at the same round; this occurs when it executes return (est i) at line 04 or 11. As observed in Lemma 2, the fact that no process decides after bt kc rounds is an immediate consequence of the code of the protocol and the round-based synchronous model. So, considering that f t processes crash, we show that no process decides after the round bf kc . Let f xk y (with y k). This means that x bf kc. The worst case scenario is when, for any process pi that evaluates the local decision predicate nbi r nbi r k, this predicate is false as many times as possible. Due to the pigeonhole principle, this occurs when exactly k processes crash during each round. This means that we have nb i n k nbi x n kx and nbi x n f n kx y , from which we conclude that r x is the first round such that nbi r nbi r y k. It follows that the processes pi that execute the round x set their can decidei boolean to true. Consequently, the processes that proceed to x decide at line 05 during that round. As x bf kc, they decide at round bf kc . T heorem
Lemma. Let H < G, Γ(G) = Γ(H) and T be an arbitrary group. Then H × T < G × T and Γ(H × T ) = Γ(G × T ). Remark. In view of Lemma the most interesting case of Xxxxxx’x question the following Question*. Let G be a simple group. Is there H < G such that Γ(G) = Γ(H)? Remark. If H < G and Γ(H) = Γ(G) then π(H) = π(G). X. X. Xxxxxxx On the finite groups Lemma. Let H < G, Γ(G) = Γ(H) and T be an arbitrary group. Then H × T < G × T and Γ(H × T ) = Γ(G × T ). Remark. In view of Lemma the most interesting case of Xxxxxx’x question the following Question*. Let G be a simple group. Is there H < G such that Γ(G) = Γ(H)? Remark. If H < G and Γ(H) = Γ(G) then π(H) = π(G). X. X. Xxxxxxx On the finite groups Lemma. Let H < G, Γ(G) = Γ(H) and T be an arbitrary group. Then H × T < G × T and Γ(H × T ) = Γ(G × T ). Remark. In view of Lemma the most interesting case of Xxxxxx’x question the following Question*. Let G be a simple group. Is there H < G such that Γ(G) = Γ(H)? Remark. If H < G and Γ(H) = Γ(G) then π(H) = π(G). X. X. Xxxxxxx On the finite groups Theorem 5 (N. M., 2014). Let G be a simple group and H < G. Then Γ(G) = Γ(H) IFF one of the following conditions holds:
Lemma. Lemma 14 The continuous clocks obtained by de nition 5 based on instantaneous clocks resulting from algorithm in gure 2 verify the precision property. More precisely: jvcn(t) vcm(t)j iv + (1+ p) tight v.
Lemma. If M is a shtuka model of E then L(M) = L(E∗, 0). Remark. Theorem 2.2 is basically the trace formula of Xxxxxxxx [2] in dis- guise. M ∈ ⊂ Remark. In general the invariant L( ) F∞ is transcendental over A F∞. Its inherent complexity reflects in the construction of the regulator making it rather involved.
Lemma. 1 The objective function Πu(Q1, Q2|ξ) defined in (6) is a jointly concave function with respect to Q1 and Q2.
Lemma. 4 is based on the orollary 1 and allows us to onstru t an unlo king fun tion from a given unlo king m/f in the ase when X is an ordered set. Let (X, ≤) be a poset and the m/f LEX ∈ P(X)X is de xxx by X LE (x) def {y ∈ X | y ≤ x}. (2.1) = Noti e that LEX ∈ UM(T). Lemma 4. Let (X, “) be a nonempty poset, P ∈ PR(X), and f ∈ UM(P ). Let G ∈ P(X)X be de xxx by X G(x) d=ef LE (x) ∩ f (x), x ∈ X, (2.2) Y d=ef {y ∈ X | G(y) /= ∅}, and the fun tion g ∈ Y Y be de xxx by ( def g(x) = ⊤G(x), ∃⊤G(x), y ∈ G(x), ¬∃⊤G(x), x ∈ Y. Then g is restri tive on (Y, “) and Fix(g) = P −1(1).
