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Discrete Mathematics with Graph Theory (3rd Edition) 20

# Discrete Mathematics with Graph Theory (3rd Edition) 20 -...

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18 Solutions to Exercises (b) We have (p - (q - {=} (p - ((-.q) Vr» {=} (-.p) V (-.q) Vr {=} (-.p) V r V (-.q) {=} -.(p /\ (-.r» V (-.q) {=} (p /\ (-.r» - (-.q). (c) (-.(p ~ q)) {=} (-.((P - q) /\ (q - pm {=} (-.(((-.p) V q) /\ ((-.q) V p))) {=} ((p /\ (-.q» V (q /\ (-.p))) {=} ((p /\ (-.q» V q) /\ ((p /\ (-.q» V (-.p)) {=} ( (p V q) /\ (( -.q) V q)) /\ ((P V (-.p)) /\ (( -.q) V (-.p))) {=} ((P V q) /\ 1) /\ (1/\ ((-.q) V (-.p))) {=} (pVq)/\((-.q)V(-.p» {=} ((-.p) V (-.q» /\ (q V p) {=} (p - (-.q» /\ ((-.q) - p) {=} (p ~ (-.q)). (d) [BB] -.[(P ~ q) V (p /\ (-.q))] {=} [-.(p ~ q) /\ -.(p /\ (-.q»] {=} [(p ~ (-.q)) /\ ((-.p) V q)], using Exercise 5(c). (e) This is an immediate application of absorption law 4(b) with p /\ (-.q) in place of p and q /\ (-.r) in place of q. (0 Using property 12 and associativity, [p - (q V r)] {=} [(-.p) V q V r] {=} [-.(p /\ (-.q))] V r (by De Morgan) {=}
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Unformatted text preview: (-.q) -r] using 12 again. (g) -.(p V q) V [(-.p) /\ q] {=} [(-.p) /\ (-.q)] V [(-.p) /\ q] (DeMorgan) {=} (-.p) /\ [(-.q) V q] (distributivity) {=} (-.p) VI{=} -'P 6. [(p/\(-.q))-q] {=} [(-.(p/\(-.q)))Vq] {=} [((-.p)Vq)Vq] {=} [(-.p)Vq]. [(P/\(-.q))-(-.p)] {=} [(-.(p/\(-.q))) V (-.p)] {=} [(-.p)VqV(-'p)] {=} [(-.p)Vq]. So these are both logically equivalent to (-.p) V q. 7. (a) We must show that A V e and 13 V e have the same truth tables, given that A and 13 have the same truth tables. This requires four rows of a truth table. A 13 e AVe 13ve T T T T T T T F T T F F T T T F F F F F The last two columns establish our claim. (b) We must show that A /\ e and 13 /\ e have the same truth tables, given that A and 13 have the same truth tables. This requires four rows of a truth table. A 13 e A /\ e 13/\e T T T T T T T F T T F F T F F F F F F F The last two columns establish our claim....
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