Mohamed G. Gouda						     CS 313K
Fall 2012							     Exercise 9

1. Let A, B, and C be three sets and let * denote the Cartesian product 
operator. Also, let (A sub B) denote the fact that A is a subset of B. Prove, 
by direct inference, that 
   	  (A sub B) => (A*C sub B*C)

Sol: 
 (x,y) in A*C
=> {Definition of *}
 (x in A) and (y in C)
=> {Definition of A sub B}
 (x in B) and (y in C)
=> {Definition of *}
 (x,y) in B*C

2. Let A and B be two sets, ^ be the intersection operator, and * be the 
Cartesian product operator. Show, by two-sided inference, that
	  (A^B)*(A^B) = (A*B)^(B*A)

Sol:
 (x,y) in (A^B)*(A^B)
<=> {Definition of *}
 (x in A^B) and (y in A^B)
<=> {Definition of ^}
 (x in A) and (x in B) and (y in A) and (y in B)
<=> {Symmetry and associativity of "and"}
 (x in A) and (y in B) and (x in B) and (y in A)
<=> {Definition of *}
 (x,y) in (A*B) and (x,y) in (B*A)
<=> {Definition of ^}
 (x,y) in (A*B)^(B*A)

3. Let A, B, and C be three sets and let v be the union operator, ^ be the 
intersection operator, and - be the set difference operator. Show, by two-sided 
inference,
	  A-(B^C) = (A-B)v(A-C)

Sol:
 x in A-(B^C)
<=> {Definition of -}
 (x in A) and (x in c(B^C))
<=> {De-Morgan's on complement "c", proven in lecture}
 (x in A) and (x in (c(B) v c(C)))
<=> {Definition of v}
 (x in A) and ((x in c(B)) or (x in c(C)))
<=> {Distribution of "and" over "or"}
 ((x in A) and (x in c(B))) or 
 ((x in A) and (x in c(C)))
<=> {Definition of -}
 (x in (A-B)) or (x in (A-C))
<=> {Definition of v}
 x in (A-B)v(A-C)

4. Let A, B, and C be three sets and let v be the union operator and * be the 
Cartesian product operator. Show, by two-sided inference, that
	  (A v B)*C = (A*C) v (B*C)

Sol:
 (x,y) in (A v B)*C
<=> {Definition of *}
 (x in A v B) and (y in C)
<=> {Definition of v}
 ((x in A) or (x in B)) and (y in C)
<=> {Distribution of "and" over "or"}
 ((x in A) and (y in C)) or 
 ((x in B) and (y in C))
<=> {Definition of *}
 (x,y) in (A*C) or
 (x,y) in (B*C)
<=> {Definition of v}
 (x,y) in (A*C) v (B*C)

5. Let A, B, and C be three sets, and let (A sub B) denotes the fact that A is 
a subset of B. Prove, by direct inference, that 
  	 (A sub B) and (B sub C) and (C sub A) 
	 => (A = B) and (B = C)

Sol:
Proving (A = B)

 (A sub B) and (B sub C) and (C sub A)
=> {From Problem 5 in Exercise 8, 
    (B sub C) and (C sub A) => (B sub A)}
 (A sub B) and (B sub A)
=> {Simple Theorem of sub}
 A = B

Proving (B = C)

 (A sub B) and (B sub C) and (C sub A)
=> {From Problem 5 in Exercise 8, 
    (C sub A) and (A sub B) => (C sub B)}
 (B sub C) and (C sub B)
=> {Simple Theorem of sub}
 B = C

6. Let PS(A) be the power set of set A. Show, by guessing values for sets A 
and B, that PS(A) v PS(B) is not necessarily equal to PS(A v B).

Sol:
Guess A = {1} and B = {2}.

PS(A) = {{}, {1}}
PS(B) = {{}, {2}}
PS(A) v PS(B) = {{}, {1}, {2}}

A v B = {1,2}
PS(A v B) = {{}, {1}, {2}, {1,2}}

Then, PS(A) v PS(B) != PS(A v B)

7. Let A, B, and C be arbitrary sets and let v denote the set union operator, 
and ^ denote the set intersection operator. Show, by two-sided inference, that
      A ^ (B v C) = (A ^ B) v (A ^ C)

Sol:
 x in A ^ (B v C)
<=> {Definition of ^}
 (x in A) and (x in (B v C))
<=> {Definition of v}
 (x in A) and ((x in B) or (x in C))
<=> {Distribution of "and" over "or"}
 ((x in A) and (x in B)) or 
 ((x in A) and (x in C))
<=> {Definition of ^}
 (x in A ^ B) or (x in A ^ C)
<=> {Definition of v}
 x in (A ^ B) v (A ^ C)

8. Let A, B, and C be three sets. Let (A sub B) denote the fact that A is a 
subset of B and "-" denote the set difference operator. Show, by contradiction, 
that 
 	(A sub B) => ((C-B) sub (C-A))

Sol:
 (Exist x, x in (C-B) and not(x in (C-A)))
=> {Definition of (C-A)}
 (Exist x, (x in (C-B)) and not(x in C and not(x in A)))
=> {De-Morgan's}
 (Exist x, (x in (C-B)) and (not(x in C) or (x in A)))
=> {Distribution of "and" over "or"}
 (Exist x, (x in (C-B)) and (not(x in C))) or		(1)
 (Exist x, (x in (C-B)) and (x in A)) 	   		(2)
 
To complete the proof, we need to prove (1)=>F and (2)=>F.

Proving (1)=>F

 (Exist x, (x in (C-B)) and (not(x in C)))
=> {Definition of (C-B)}
 (Exist x, (x in C) and not(x in B) and not(x in C))
=> {(x in C) and not(x in C) => F}
 F

Proving (2)=>F
 (Exist x, (x in (C-B)) and (x in A))
=> {Definition of C-B}
 (Exist x, (x in C) and not(x in B) and (x in A))
=> {A sub B}
 (Exist x, (x in C) and not(x in B) and (x in B))
=> {not(x in B) and (x in B) => F}
 F