Mohamed G. Gouda						     CS 313K
Summer 2013							     Homework 1

1) Let x and y be two parameters whose values are in the set D = {a,b}.
Define two predicates P(x,y) and Q(x,y) such that 
	(Exist x, y, P(x,y) and Q(x,y)) !=
	(Exist x, y, P(x,y)) and (Exist x, y, Q(x,y))
Explain your answer.

Sol:
Define P(x,y) and Q(x,y) as follows:

P(a,a)=T  P(a,b)=F  P(b,a)=T  P(b,b)=F
Q(a,a)=F  Q(a,b)=T  Q(b,a)=F  Q(b,b)=T

Explanation:
(Exist x, y, P(x,y) and Q(x,y)) = F
(Exist x, y, P(x,y)) = T
(Exist x, y, Q(x,y)) = T
(Exist x, y, P(x,y)) and (Exist x, y, Q(x,y)) = T

Thus, 
(Exist x, y, P(x,y) and Q(x,y)) != 
(Exist x, y, P(x,y)) and (Exist x, y, Q(x,y))

2) Let x be a parameter whose value is in the set Dx={a,b}, and let y
be a parameter whose value is in the set Dy={1,2,3}. 

(a) Define a predicate P(x,y) such that
(All v in Dx, Exist w in Dy, P(v,w)) = (Exist w in Dy, All v in Dx, P(v,w))

(b) Define a predicate Q(x,y) such that
(All v in Dx, Exist w in Dy, Q(v,w)) != (Exist w in Dy, All v in Dx, Q(v,w))

Explain your answers.

Sol:
(a)
P(a,1)=F  P(a,2)=T  P(a,3)=F
P(b,1)=T  P(b,2)=T  P(b,3)=F

Explanation:
(All v in Dx, Exist w in Dy, P(v,w)) = T
(Exist w in Dy, All v in Dx, P(v,w)) = T

Thus,
(All v in Dx, Exist w in Dy, P(v,w)) = (Exist w in Dy, All v in Dx, P(v,w))

(b)
Q(a,1)=F  Q(a,2)=T  Q(a,3)=F
Q(b,1)=T  Q(b,2)=F  Q(b,3)=T

Explanation:
(All v in Dx, Exist w in Dy, Q(v,w)) = T
(Exist w in Dy, All v in Dx, Q(v,w)) = F

Thus,
(All v in Dx, Exist w in Dy, Q(v,w)) != (Exist w in Dy, All v in Dx, Q(v,w))

3) Prove by contradiction the predicate
	((x is odd) and (y is odd)) => (x*y is odd)

Sol:
	(x is odd) and (y is odd) and not(x*y is odd)
=> {integer is either odd or even}
	(x is odd) and (y is odd) and (x*y is even)
=> {definitions of odd and of even}
	(x = 2m+1) and (y = 2n+1) and (x*y = 2k)  for integers m, n, k
=> {compute x*y}
	(x*y = 2(2mn + (m+n)) + 1 and (x*y = 2k) for integers m, n, k
=> {2mn + (m+n) is an integer r}
	(x*y = (2r+1)) and (x*y = 2k) for integers r, k
=> {arithmetics}
	2r+1 = 2k for integers r, k
=> {arithmetics}
	(k-r) = 1/2 for integers r, k
=> {definition of integers subtraction}
	F

4) Prove by indirect inference the following predicate: 
	(x*y is odd) => ((x is odd) and (y is odd))

Sol:
	not (x is odd) or not (y is odd)
=> {an integer is either odd or even}
	(x is even) or (y is even)
=> {an integer is either odd or even}
	(x is even) and (y is even or y is odd)
=> {distribution of and over or}
	(x is even and y is even) or 
	(x is even and y is odd)
=> {definitions of odd and of even}
	(x=2m and y=2n) or 
	(x=2m and y=2n+1)  for integers m, n
=> {compute x*y}
	(x*y=2r) or
	(x*y=2r)  for integer r
=> {definition of even}
	(x*y is even)
=> {an integer is either odd or even}
	not (x*y is odd)