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# Discrete Mathematics and Its Applications with MathZone

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ICS 6D ( Discrete Mathematic): Solution to Homework #1 Section 1.1, problem 62 Let’s denote: K = “Kevin is chatting” A = “Abby is chattnig” H = “Heather is chatting” R = “Randy is chattnig” V = “Vijay is chatting” Here’s what we know about the world (in the order as they are stated in this exercise): 1. K H 2. R V 3. A R 4. ( V K ) ( ¬ V ∧ ¬ K ) 5. H ( A K ) So how do we know which combination is true? We have five variables here and 5 logical formulas which impose relations between these variables. So what we can do is to do a series of transformations which would simplify these relations: 2’. We can replace (2) by equation R = ¬ V . 3’. Therefore we can replace (3) by A → ¬ V . 4’. Also we can replace (4) by an equivalent expression V K Now let’s pick some variable on which lots of other variables seem to depend. For example pick H , set it to true, and see what happens to the other variables: If H then by (5) we immediately get A and K . (1) is true. By (3’) from A we get ¬ V . However, by (4’)
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Unformatted text preview: From ¬ V we get ¬ K . But that’s a contradition , because we cannot have both K and ¬ K . What does this mean? Well, it means that H cannot be true, because that assumption led to a contradi-tion which violated the set oF Facts 1-5 we were told about the world. (In particular it led to a contradictory statement that K ∧¬ K . ThereFore we can conclude that ¬ H . Assuming ¬ H , by (1) we get K . By (4’) we get V . By (3’) we get ¬ A . By (2’) we get ¬ R . And the only constraint leFt, (5), is satisfed because ¬ H . This leads us to a solution ( K, ¬ A, ¬ H, ¬ R,V ) which, you can check, does satisFy all the criteria 1-5. It is also a unique solution because we showed two things: ±irst, that H cannot happen. Second, that iF ¬ H then the other Four variables have uniqely assigned values, which are all a consequence oF ¬ H and Facts 1-5. ThereFore this is also the only solution. Page 1 oF 1...
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