1) Consider bit strings of 3 bits b2 b1 b0. Develop a proposition R expressed in terms of b0, b1, and b2 which is true (T) for such a bit string if and only if b2 b1 b0 = b0 b1 b2, i.e., if the bit string b2 b1 b0 is equal to its reverse. For example, R should be true (T) for bit string 010, but false F for bit string 001. Simplify you proposition as much as possible using the replacements laws of propositional logic. 2) Suppose that you have 2 two-bit bit strings a and b. Think of a as a1a0 and b as b1b0. So a(and, similarly b) represent the natural numbers 0, 1, 2, and 3 as the bit strings 00,01, 10, and 11, respectively. Design a proposition L in terms of a0, a1, b0, and b1 such that L is true if and only if a b. For example, L should be true (T) when a = 10 and b = 11, L should be false when a= 01 and b = 00, and L should also be true when a = 01 and b = 01. Simplify you proposition as much as possible using the replacements laws of propositional logic. 3) Suppose that you have 2 two-bit bit strings a and b. Think of a as a1a0 and b as b1b0. So a and b represent the numbers 0, 1, 2, and 3 as the bit strings 00, 01, 10, and 11, respectively. Design 2 propositions in terms of a0, a1, b0, and b1 that "implement" the function max (0, a-b). So for example max(0,3-0) = 3 and max(0,1-2) = 0. Notice that you need 2 propositions since max ranges from 0 to 3. Simplify your propositions as much as possible using the replacement laws.