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# Discrete Math EECS 203

UM

GPA 3.8

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This 5 page Class Notes was uploaded by Ophelia Ritchie on Thursday October 29, 2015. The Class Notes belongs to EECS 203 at University of Michigan taught by Benjamin Kuipers in Fall. Since its upload, it has received 156 views. For similar materials see /class/231521/eecs-203-university-of-michigan in Engineering Computer Science at University of Michigan.

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Date Created: 10/29/15

Discussion Notes 17 EECS 203 1 Section 11 14 Propositions Let p7 q7 and r be propositions o p You get an A on the nal exam 0 q You do every exercise in the textbook 0 7 You get an A in the class Write these propositions using pg and r and logical connectives 7 a etc 0 You get an A in this class7 but you do not do every exercise in the book 7 You get an A in this class is 7 You do every exercise in the textbook is g you do not do every exercise in the book is q When you combine the rst part of the sentence with the second part7 it tells you r g 0 To get an A in this class7 it is necessary for you to get an A on the nal 7 get an A in this class is 7 get an A on the nal is p Note that the sentence doesn7t say anything about what else might be important Perhaps it is the case that you need to do all of the exercises as well Perhaps you get an A on the nal7 but you still dont pass All the statement above says is that an A in the class irnplies you got an A on the nal7 or else you couldnt have gotten the A in the class Therefore7 r a p is how you write the above statement 0 You will get an A in this class if and only if you either do every exercise in the textbook or you get an A on the nal 7 You will get an A in this class is 7 do every exercise in the textbook is q you get an A on the nal is p H and only if is lt gt After the if and only if 7 we have q or p This is from either do every exercise in the textbook or you get an A on the nal Connecting the pieces gives 7 lt gt q Vp 2 Sixth Edition Section 11 23 Conditional Statements State the converse inverse and contrapositive of the following statements o If it snows tonight then I will stay at home p it snows tonight q I will stay at home lnverse If NOT p then NOT g If it does not snow tonight then I will not stay at home Converse If Q then p If I stay at home tonight then it will snow Contrapositive If NOT q then NOT p If I do not stay at home tonight it will not have snowed o I go to the beach whenever it is a sunny summer day First it helps to restate in the familiar form lfp then q If it is a sunny summer day then I go to the beach p it is a sunny summer day q I go to the beach lnverse lf NOT p then NOT Q If it is not a sunny summer day then I do not go to the beach Converse If Q then p If I go to the beach then it is a sunny summer day Contrapositive lf NOT q then NOT p lfl do not go to the beach then it is not a sunny summer day 0 When I stay up late it is necessary that I sleep until noon Restating in if p then q form gives lfl stay up late then I sleep until noon p I stay up late q I sleep till noon lnverse lf NOT p then NOT q lfl do not stay up late then I will not sleep until noon Converse If Q then p If I sleep until noon I stayed up late Contrapositive lf NOT q then NOT p If I do not sleep until noon then I did not stay up late 3 Sixth Edition Section 11 28 cdf Compound Propositions Make truth tables for the following propositions O 0 p69 pV q First enumerate all combinations of the variables p and q This gives the rst and second columns Then consider qu This is only false when both p and q are false giving the third column Finally the exclusive or requires that only one half77 of the statement be true This means that the fourth column will display true on rows where the rst and third column contain one true and one false That is only the case on row three meaning that the entire statement is true for false p and true q Otherwise as in the rest of the fourth column the statement evaluates to false wmaas maniac mawm Aaanw For this problem the left and right sides are pursued as before giving columns three and four For the a think on when the statement If a then b77 can be shown to be false If a and b are both true the statement that a a b is not false If a is false b can still be true so a a b is not false If both are false the statement a a b is not false However if a is true but b is false then clearly 1 cannot imply b To get the fth column apply this logic to columns three and four Aw Vw WWQQB WQWQQ gt T T T T T T F T 4 Section 11 44 ab Bitwise Operations Evaluate each of these expressions 0 11000 01011 V 11011 Focus on the right 01011 V 11011 The rst bit on the left is a zero and the rst bit on the right is a one This corresponds to F V T7 which is true This means that the rst bit in the evaluation of 01011 V 11011 is a 1 We continue through the rest of the bits7 evaluating 01011 V 11011 as 11011 Now we refer back to the original expression and replace the right 11000 11011 Here7 we are dealing with 77and The rst digit of both the left and the right sides is 1 T T is true7 meaning that the rst digit of the evaluation is also a 1 This gives 11000 11011 as 11000 01111 10101 V 01000 Starting with the rst part 0 01111 10101 00101 This result 00101 can now be put back into the original expression This leaves 00101 V 01000 Once again7 noting that this is an or operation7 proceed bit by bit The rst digit is 0 V 07 which is 0 The second digit is 0 V 17 which is 1 The nal result is 01101 5 Section 1 2 42 Logic Circuits Construct a combinatorial circuit using inverters OR gates and AND gates that represents pA T v hm Construct in parts The rst part needs an and a A We can just add a negation to r then AND that output with p However we will also need a normal nonrnegated r for the second expression So for the second expression before the negation of r we branch off so that we can AND r with a negated q The left expression in parentheses is then formed by the output of the top AND he right expression is formed by the output of the lower AND A11 that is left is to OR the left expression with the right expression Thus the OR gate takes as its inputs the outputs of the two AND gates The total Circuit s output out of the OR forms the total expression D AND NOT r rAND NOT q

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