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Mohamed G. Gouda						     CS 311
Fall 2014							     Midterm 3
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1. (5 points)
Let A be any set. Use direct inference to prove the predicate:
	(x in A) => ({x} sub A)
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2. (5 points)
Let f:A-->B and g:B-->C be two functions. Use direct inference to prove the
predicate:
	((g.f) is injective)  =>  (f is injective)
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3. (5 points)
Let A be the set of non-negative integers. Also let f: A -> A be the function
	(All x in A, f(x) = x+1)
Write down an argument that shows function f has no inverse.
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4. (5 points)

Consider the following predicate:
	(f(n) is O(g(n))) <=> (f(n) is O(3*g(n)))

A two-sided inference proof of this predicate is as follows.

	(f(n) is O(g(n)))
<=> {hint 1}
	(Exist positive real numbers K and C,
		(All integer n, n>K, |f(n)| =< C*|g(n)|) )
<=> {hint 2} 
	(Exist positive real numbers K and C,
		(All integer n, n>K, |f(n)| =< (C/3)*|3*g(n)|) )

<=> {hint 3}
	(Exist positive real numbers K and C/3,
		(All integer n, n>K, |f(n)| =< (C/3)*|3*g(n)|) )

<=> {hint 4}
	(f(n) is O(3*g(n)))

Write down the four missing hints in this proof.
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Solutions
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1.
	y in {x}
=> {definition of {x}}
	y = x
=> {x in A}
	y in A
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2.
	f(x1) = f(x2)
=> {g is a function}
	g(f(x1)) = g(f(x2))
=> {definition of "."}
	(g.f)(x1) = (g.f)(x2)
=> {g.f is injective}
	x1 = x2
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3.
We show that f is not surjective and so it is not bijective and has no 
inverse.

From definition of f, we have
	(Exist y in A, y = 0, not(Exist x in A, x+1 = y))

Thus f is not surjective
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4.
hint 1 : 	defintion of f(n) being O(g(n))
hint 2 : 	C*|g(n)| = (C/3)*|3*g(n)|
hint 3 : 	(Exist positive reat number C, P(C/3)) <=> 
		(Exist Positive real number C/3, P(C/3)) 
hint 4 : 	definition of f(n) being O(3*g(n))
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