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Discrete Mathematics with Graph Theory (3rd Edition) 24

Discrete Mathematics with Graph Theory (3rd Edition) 24 - p...

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22 Solutions to Exercises (e) This argument is valid. The second and third premises give q by Modus Ponens. Together with the first premise (double negation and dijunctive syllogism), we get .p. (t) This argument is valid. The first and third premises give .p by modus tollens. Together with the second premise (double negation and a second application of modus tollens), we get r. 2. (a) If p is true and p implies q, then q is true. (b) This was PAUSE 5. (c) If p is true or q is true and p is not true, then q must be true. (d) We do this by contradiction. So assume the premises are true but the conclusion is false. Since p ~ r is false, we must have p true and r false. Since q ~ r is true and r is false, q must be false. Then, since p ~ q is true, p must be false, giving a contradiction. (e) We prove this by contradiction. Assume p V qis false. This means both p and q are false. Since p V r is true and
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Unformatted text preview: p is false, r must be true. Since q V (.r) is true and q is false, r must be false, giving a contradiction. 3. (a) [BB] We analyze with a truth table. There are five rows when the premises are all true and in each case the conclusion is also true. The argument is valid. (b) We analyze with a partial truth table showing the nine situations in which both the premises are true. In every case, the conclusion is true. The argument is valid. p T T F F T T F F p T T T F F F F F F q r T T F T T T F T T F F F T F F F r q T T T F T F T T T F T F F T F F F F pVq p-+r T T T T T T F T T F T F T T F T s pAq T T T F F F T F T F F F T F T F F F q-+r (PV q) -+ r T T T T T T T T F F T F F F T T rAs (pAq) -+ (rAs) T T T T F T T T T T F T F T F T F T * * * * * (c) The second and third premises are p ~ rand r ~ s which together imply p ~ s by the chain rule. Thus the argument becomes pVq p~s qVs...
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