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Mohamed G. Gouda						     CS 311
Fall 2015							     Midterm 1
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50 minutes
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1. (5 points)
Are there Boolean values x and y that make the following formula true
	(not y) = ((not x  =  y) xor (x -> y))
Explain your answer.
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2. (5 points)
Let Dx be the set {a, b, c, d} and let P(x) be the predicate defined as follows:
	P(a) = F
	P(b) = T
	P(c) = F
	P(d) = T
What are the Boolean values of the following predicates:
	1. (All u in Dx, P(u)) -> (Exist u in Dx, P(u))
	2. (P(a)) -> (Exist u in Dx, P(u))
Explain your answers.
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3. (5 points)
Definition: Let u and v be positive integers. Integer u is said to be bigger than 
integer v, denoted u >> v, iff there is a positive integer m such that u = (v+m).

Let Dx, Dy, and Dz be the set of all positive integers. Use direct inference to
prove the following predicate:
	(All x, y, z, ((x >> y) and (y >> z)) => (x >> z)) 
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4. (5 points)
Refer to the definition of >> in Problem 3.

Let Dx and Dy be the set of all positive integers. Use indirect inference to
prove the following predicate:
	(All x, y, (x >> y) => (not (x=y)))

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Solutions
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1.
x	y	not y	(not x)=y	x->y	xor	=

F	F	T	F		T	T	F
F	T	F	T		T	F	T
T	F	T	T		F	T	T
T	T	F	F		T	T	T

Thus, x=T and y=T will make the predicate T
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2.
	1.	(All u in Dx, P(u))  =	F
		Thus, the predicate is (F -> anything) = T
	2.	(P(a)) = F
		Thus, the predicate is (F -> anything) = T
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3.
	(x >> y) and (y >> z)
=> {definition of >>}
	(x = y+m) and (y = z+n) for some positive integers m and n
=> {replace y by z+n}
	x = z+m+n for some positive integers m and n
=> {m+n is a positive integer r}
	x = z+r for some positive integer r
=> {definition of >>}
	x >> z
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4.
	not(not (x=y))
=> {idempotence}
	x = y
=> {arithmetics}
	x =! y+m for every positive integer m
=> {definition of not(x >> y)}
	not(x >> y)
------------------------------------------------------------------------------
Mohamed G. Gouda						     CS 311
Fall 2015							     Midterm 1
------------------------------------------------------------------------------
50 minutes
------------------------------------------------------------------------------

1. (5 points)
Are there Boolean values x and y that make the following formula true
	(not y) = ((x  =  not y) xor (x -> y))
Explain your answer.
-------------------------------------------------------------------------------
2. (5 points)
Let Dx be the set {a, b, c, d} and let P(x) be the predicate defined as follows:
	P(a) = T
	P(b) = F
	P(c) = T
	P(d) = F
What are the Boolean values of the following predicates:
	1. (All u in Dx, P(u)) -> (Exist u in Dx, P(u))
	2. (P(a)) -> (Exist u in Dx, P(u))
Explain your answers.
--------------------------------------------------------------------------------
3. (5 points)
Definition: Let u and v be positive integers. Integer u is said to be bigger than 
integer v, denoted u >> v, iff there is a positive integer m such that u = (v+m).

Let Dx, Dy, and Dz be the set of all positive integers. Use direct inference to
prove the following predicate:
	(All x, y, z, ((x >> y) and (y >> z)) => (x >> z)) 
---------------------------------------------------------------------------------
4. (5 points)
Refer to the definition of >> in Problem 3.

Let Dx and Dy be the set of all positive integers. Use indirect inference to
prove the following predicate:
	(All x, y, (x >> y) => (not (x=y)))

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Solutions
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1.
x	y	not y	(x = not y)	x->y	xor	=

F	F	T	F		T	T	T
F	T	F	T		T	F	T
T	F	T	T		F	T	T
T	T	F	F		T	T	F

Thus, x=F and y=F will make the predicate T
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2.
	1.	(All u in Dx, P(u))  =	F
		Thus, the predicate is (F -> anything) = T
	2.	(P(a)) = T
		(Exist u in Dx, P(u)) = T
		Thus, the predicate is (T -> T) = T
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3.
	(x >> y) and (y >> z)
=> {definition of >>}
	(x = y+m) and (y = z+n) for some positive integers m and n
=> {replace y by z+n}
	x = z+m+n for some positive integers m and n
=> {m+n is a positive integer r}
	x = z+r for some positive integer r
=> {definition of >>}
	x >> z
----------------------------------------------------------------------------------
4.
	not(not (x=y))
=> {idempotence}
	x = y
=> {arithmetics}
	x =! y+m for every positive integer m
=> {definition of not(x >> y)}
	not(x >> y)
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