Common use of Aside Clause in Contracts

Aside. non-associative scheme‌ A non-associative key agreement scheme is any scheme k that is not asso- ciative. / In particular, Diffie–▇▇▇▇▇▇▇ key agreement k (Definition 2.2.1) is not as- sociative. To see this, first k1(a, b) = k3(a, b) = ba = ab = k2(b, a) = k4(b, a), for most a and b. But if k were associative, then it would be the subscheme of a multiplicative scheme K, and would have k1(a, b) = K1(a, b) = K2(a, b) = k2(a, b), a contradiction. − − ∈ { } For a smaller non-associative scheme, consider rock, scissors, paper key agreement again. Represent it as k1(a, b) = a and k2(b, c) = c and k3(a, b) = k4(a, b) = (a b) mod 3, where all a, b, c Z/3 = 0, 1, 2 , making each ki a binary operation on Z/3. Recall that this is a key agreement scheme because k3(a, k2(b, c)) = k3(a, c) = (a c) mod 3 = k4(a, c) = k4(k1(a, b), c). Suppose that k is a subscheme of a multiplicative scheme K. Then all the sessions of k must be sessions of K, so k1(a, b) = K1(a, b) and k2(b, c) = K2(b, c). But since K is multiplicative, K is symbiotic with K1 = K2. Clearly, k1 k cannot be a subscheme of K. k2, so

Aside. non-associative scheme‌ A non-associative key agreement scheme is any scheme k that is not asso- ciative. / ƒ In particular, Diffie–▇▇▇▇▇▇▇ key agreement k (Definition 2.2.1) is not as- sociative. To see this, first k1(a, b) = k3(a, b) = ba = ab = k2(b, a) = k4(b, a), for most a and b. But if k were associative, then it would be the subscheme of a multiplicative scheme K, and would have k1(a, b) = K1(a, b) = K2(a, b) = k2(a, b), a contradiction. − − ∈ { } For a smaller non-associative scheme, consider rock, scissors, paper key agreement again. Represent it as k1(a, b) = a and k2(b, c) = c and k3(a, b) = k4(a, b) = (a b) mod 3, where all a, b, c Z/3 = 0, 1, 2 , making each ki a binary operation on Z/3. Recall that this is a key agreement scheme because k3(a, k2(b, c)) = k3(a, c) = (a c) mod 3 = k4(a, c) = k4(k1(a, b), c). Suppose that k is a subscheme of a multiplicative scheme K. Then all the sessions of k must be sessions of K, so k1(a, b) = K1(a, b) and k2(b, c) = K2(b, c). But since K is multiplicative, K is symbiotic with K1 = K2. Clearly, k1 k cannot be a subscheme of K. k2, so