can you just do lst part with the dont care's. i an confused with how the table works

7) (Synthesis of a decomposed Boolean function by exploiting don't care conditions - 20 points) Consider the Boolean function F(a, b, c, d, e) whose K-map is shown below in Fig. 2 (i). Now suppose that a logic synthesis algorithm decomposes Fas F(a, b, c, d,e) = (o(a,b,c). (a,b,c),d,e), shown in Fig. 2, where the SOP representations are: Functions 90 = a'be + abc + abd' and 91 a'c + abe Function h = good' +9091 d' + %91e You are asked to solve the following: a) From the K-map of F(a, b, c, d, e), identify a minimum SOP form representation of F in its undecom- posed form in terms of the primary inputs {a,b,c,d,e). What is the SOP literal cost of F? b) Now assume the a decomposition is applied as shown in Fig. 2. Minimize the SOP form of 90.91, h. Are they already given in minimal form? c) This decomposition creates don't care conditions at the input of the h(90.91.d,e) block. Identify the don't care conditions at the input of h. d) Using the don't care conditions, minimize the SOP form of h. What is the total SOP literal cost of 90,91 and h. Do the don't care conditions result in further logic simplification with literal cost savings? a 0 0 0 0 1 1 1 1 b 0 0 1 1 1 1 0 0 c011 1.0, 0110 ooo 11 on 1 de 01 0 1 1 0 1 1 1 0 00 00 10000 10001 (i) F(a, b, c, d, e) don't care conditions? (ii) A decomposed implementation of F(a,b,c,d,e) Fig. 2. Decomposition of F(a, b, c, d,e) = n(90(a,b,c), 9 (a, b, c), d,e). Compute the don't cares at the input of the h(90,91, d, e) block and simplify the SOP form of h