Chapter 1 - Section 1.1 - Propositional Logic - Exercises - Page 16: 44

44. Evaluate each of these expressions. a) 1 1000 $\land$ (0 1011 $\vee$ 1 1011) Answer: 1 1000 b) (0 1111 $\land$ 1 0101) $\vee$ 0 1000 Answer: 0 1101 c) (0 1010 $\oplus$ 1 1011) $\oplus$ 0 1000 Answer: 1 1001 d) (1 1011 $\vee$ 0 1010) $\land$ (1 0001 $\vee$ 1 1011) Answer: 1 1011

There are three operations involved in this problem $\land$(AND), $\vee$(OR), and $\oplus$(XOR) AND is only true when both propositions are true, so false $\land$ false = false false $\land$ true = false true $\land$ false = false true $\land$ true = true OR is true when either, or both of the propositions are true, so false $\vee$ false = false false $\vee$ true = true true $\vee$ false = true true $\vee$ true = true XOR(also known as Exclusive Or) is true when, either, but not both of the propositions are true, so false $\oplus$ false = false false $\oplus$ true = true true $\oplus$ false = true true $\oplus$ true = false The result of these operations remain the same when we substitute 0 for false, and 1 for true. When we then extend bits into bit strings, we can then apply bitwise OR, bitwise AND, and bitwise XOR. So for part a) 1 1000 $\land$ (0 1011 $\vee$ 1 1011), we must first do the bitwise operation 0 1011 $\vee$ 1 1011 to do so, we line up the bits 0 1011 1 1011 and apply $\vee$ to each vertical pair of bits, resulting in 1 1011 which we can then use to finish the evaluation of the expression 1 1000 $\land$ 1 1011 = 1 1000, which is the result of evaluating the whole expression Our next steps for the following problems are apply similarly b) (0 1111 $\land$ 1 0101) $\vee$ 0 1000 0 1111 $\land$ 1 0101 = 0 0101 0 0101 $\vee$ 0 1000 = 0 1101 (The Answer) c) (0 1010 $\oplus$ 1 1011) $\oplus$ 0 1000 0 1010 $\oplus$ 1 1011 = 1 0001 1 0001 $\oplus$ 0 1000 = 1 1001 (The Answer) d) (1 1011 $\vee$ 0 1010) $\land$ (1 0001 $\vee$ 1 1011) 1 1011 $\vee$ 0 1010 = 1 1011 1 0001 $\vee$ 1 1011 = 1 1011 (1 1011 $\vee$ 0 1010) $\land$ (1 0001 $\vee$ 1 1011) = 1 1011 $\land$ 1 1011 1 1011 $\land$ 1 1011 = 1 1011 (The Answer)