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Mohamed G. Gouda						CS 311 
Fall 2014							Homework 3
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1. (2 points) 
Let A be the set that has the three elements { }, 1, and {2, 3}. What are the 
eight elements of set PS(A) of set A. 
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2. (2 points)
(a) Use direct inference to prove the following predicate true 
	(A sub (B^C)) => (A sub B)
where A, B, and C are sets, "sub" denotes "subset", and "^" denotes the set
intersection operator. 

(b) Use guessing to prove the following predicate false
	(A sub (BvC)) => (A sub B)
where A, B, and C are sets, "sub" denotes "subset", and "v" denotes the set
union operator.
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3. (2 points)
Use direct inference to prove the predicate 
	(A sub B) => ((A^B sub A) and (A sub A^B))
where A and B are sets, "sub" denotes "subset", and "^" denotes the set
intersection operator. 

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4. (4 points)
Use by-contradiction to prove the predicate
	(A sub B) => (A-B = { })
where A and B are sets, "sub" denotes "subset", and "-" denotes the set
difference operator.

(You can use the following hint in your proof:
	{((Exist x, P(x)) and (All y, not(P(y)))) => F} )
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Solutions
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1. The eight elements of set PS(A) are:
	{ }, {{ }}, {1}, {{2, 3}}, {{ }, 1}, {{ }, {2, 3}}, {1, {2, 3}},
	{{ }, 1, {2, 3}}
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2.
(a)	
	x in A
=> {A sub B^C}
	x in B^C
=> {definition of ^}
	(x in B) and (x in C)
=> {(P and Q) => P}
	x in B

(b)
Let A be {0}, B be {1}, and C be {0}. Thus, BvC = {0, 1} and so (A sub BvC) but
not(A sub B).
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(3)
First, prove (A sub B) => (A^B sub A)
	x in A^B
=> {definition of ^}
	(x in A) and (x in B)
=> {(P and Q) => P}
	x in A

Second, prove (A sub B) => (A sub A^B)
	x in A
=> {A sub B}
	(x in A) and (x in B)
=> {definition of ^}
	x in (A^B)
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(4)
	not (A-B = { })
=> {definition of { } set)
	Exist x, x in (A-B)
=> {definition of A-B}
	Exist x, (x in A) and not(x in B)
=> {definition of A sub B}
	(Exist x, (x in A) and not(x in B)) and
	(All y, not(y in A) or (y in B))
=> {((Exist x, P(x)) and (All y, not(P(y)))) => F}
	F
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