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Mohamed G. Gouda						    CS 311
Fall 2015							    Homework 1
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1. Use equivalence laws to prove that the following two formulas are equivalent:
	f(x) = not(f(x))
	F
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2. Let Dx be the set {1,2} and Dy be the set {1,2,3}. Define the parametrized
predicate P(x,y) such that the following predicate is T:
	(All x, Exist y, P(x,y)) and (not(Exist y, All x, P(x,y)))
Explain your answer.
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3. A positive integer x is said to be divisble by a positive integer m, denoted
x|m, iff there is a positive integer k such that x = (m*k) where * denotes the
integer multiplication operator. Let Dx, Dy, and Dm be the set of all positive 
integers. Use direct inference to prove the following predicate is T:
	(All x, y, m, (x|m and y|m) => ((x+y)|m))
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4. Give a by-contradiction proof of the predicate:
	(7 is even) => (7*3 is even))
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Solutions
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1.
	f(x) = not(f(x))
=> {definition of the Bolean operator "="}
	(f(x) and not(f(x))) or (not(f(x)) and (not(not(f(x)))))
=> {idempotence}
	(f(x) and not(f(x))) or (not(f(x)) and (f(x)))
=> {symmetry of "and"}
	(f(x) and not(f(x))) or ((f(x)) and (not(f(x))))
=> {idempotence}
	f(x) and not(f(x))
=> {elementary}
	F
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2. Define P(x,y) as follows:
	P(1,1) = T
	P(1,2) = F
	P(1,3) = F
	P(2,1) = F
	P(2,2) = T
	P(2,3) = F
In this case, we have
	(All x, Exist y, P(x,y)) = T 
	(Exist y, All x, P(x,y)) = F
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3.
	(x|m and y|m) 
=> {definition of "|"}
	x = (m*k) for some positive integer k and
	y = (m*r) for some positive integer r
=> {adding x plus y}
	x+y = m*(k+r) for some positive integers k and r
=> {(k+r) is a positive integer s}
	x+y = m*s for some positive integer s
=> {definition of "|"}
	((x+y)|m))
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4.
	((7 is even) and not(7*3 is even))
=> {7 is not even}
	((7 is even) and not(7*3 is even)) and not(7 is even)
=> {symmetry and associativity of and}
	((7 is even) and not(7 is even)) and not(7*3 is even)
=> {idempotence}
	(F and not(7*3 is even))
=> {elementary}
	F
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