Just need truth table, and boolean expression (simplified) for these thumbs up for answering both or even just #3

2. (2 points) Given four inputs: a, b, c & d, where (a, b) represents a 2-bit unsigned binary number X; and (c,d) represents a 2-bit unsigned binary number Y (i.e. both X and Y are in the range #0 to #3). The output is z, which is 1 whenever X >Y, and 0 otherwise (this circuit is part of a "2-bit comparator"). For instance, if a = 1, b = 0 (i.e. X = b10 => #2); c = 0, d = 1 (i.e. Y = b01 => #1); then z = 1, since b10 > b01 3. (2 points) Given four inputs: a, b, c&d, where (a, b) represents a 2-bit unsigned binary number X; and (c,d) represents a 2-bit unsigned binary number Y (i.e. both X and Y are in the range #0 to #3) - i.e. same as q. 2 This time, there are 4 outputs - i.e. in effect, you are building 4 separate truth tables -03, 02, 01, 00. These correspond to the unsigned 4-bit product of the two numbers X, Y E.g., if a = 1, b = 0 (i.e. X = b10 => #2); c = 1, d = 1 (i.e. Y = 511 => #3); then {03, 02, 01, 00} = {0, 1, 1, 0}, since #2 * #3 = #6. Derive the boolean expressions only for columns 00 and 02