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# 1 MATH 135 F 2012: Assignment 2 Due:30 AM, Wed., 2012 Sep. 26 in the dropboxes outside MC 4066 or in the electronic dropbox for students in the...

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1 MATH 135 F 2012: Assignment 2 Due: 8:30 AM, Wed., 2012 Sep. 26 in the dropboxes outside MC 4066 or in the electronic dropbox for students in the online section Write your answers in the space provided. If you wish to typeset your solutions, use one of the solution templates posted on the course web site. Family Name: First Name: I.D. Number: Section: Mark: (For the marker only.) If you used any references beyond the course text and lectures (such as other texts, discussions with colleagues or online resources), indicate this information in the space below. If you did not use any aids, state this in the space provided. 1. A tautology is a logical statement that is true for all possible values of the component statements. Use a truth table to determine whether or not ( A ( B C )) ⇐⇒ (( A ∧ ¬ B ) C ) is a tautology. A B C T T T T T F T F T T F F F T T F T F F F T F F F
2 2. Are the following pieces of reasoning valid? (Does the third statement follow from the ﬁrst two?) Explain. (a) If I do every problem in the text book, then I will learn discrete mathematics. I learnt discrete mathematics. Therefore, I must have done every problem in the text book. (b) If I do every problem in the text book, then I will learn discrete mathematics. I did not learn discrete mathematics. Therefore, there is a problem in the book which I did not do. (c) If I miss a lecture, then I am ill or I had a late night. I had a late night. Therefore, I missed a lecture. 3. An integer p > 1 is called a prime if its only divisors are 1 and p . Otherwise, p is called composite . Consider the proposition below. Proposition 1. For every prime number p , p + 7 is composite. (a) Identify the four parts of the quantiﬁed statement. Quantiﬁer: Variable: Domain: Open sentence: (b) Prove the proposition.
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Q1: Let A ⇒ (B ∨ C ) be P
and (A ∧ ¬B ) ⇒ C be Q . 1
A
F
F
F
F
T
T
T
T 2
B
F
F
T
T
F
F
T
T 3
C
F
T
F
T
F
T
F
T 4
B∨C
F
T
T
T
F
T
T
T 5
A ⇒ (B ∨ C )
T
T
T
T
F
T
T
T 6
¬B
T
T
F
F
T
T...

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