Mohamed G. Gouda						     CS 313K
Fall 2012							     Midterm 4

1. Let A and B be arbitrary sets, and let "v" denote the set union operator,  
"^" denote the set intersection operator, and "not" denote the set complement 
operator. Show, by two-sided inference, that 
  	     not(A ^ B) = not(A) v not(B)

Sol:
 x in not(A ^ B)
<=> {Definition of "not"}
 not(x in (A ^ B))
<=> {Definition of "^"}
 not((x in A) and (x in B))
<=> {De-Morgan's}
 not(x in A) or not(x in B)
<=> {Definition of "not"}
 (x in not(A)) or (x in not(B))
<=> {Definition of "v"}
 x in (not(A) v not(B))

2. Let f:A->B and g:B->C be two functions. Show, by guessing values for A, B, C,
f, and g, that the following inference does not hold: 
       (g.f is surjective) => (f is surjective)

Sol:
	A={1}, B={2,3}, C={4}
	f(1)=2, 
	g(2)=4, g(3)=4, 
	g.f(1)=g(2)=4
Thus, g.f is surjective but f is not surjective

3. Let f:A->B be a function, and let A1 and A2 be two non-empty subsets of A. 
Show, by two-sided inference, that
	  f(A1) v f(A2) = f(A1 v A2)

Sol:
 y in (f(A1) v f(A2))
<=> {Definition of v}
 (y in f(A1)) or (y in f(A2))
<=> {Definitions of f(A1) and f(A2)}
 (Exist x in A1, f(x) = y) or 
 (Exist x in A2, f(x) = y)
<=> {Definition of v}
 (Exist x in A1 v A2, f(x) = y) 
<=> {Definition of f(A1 v A2)}
 y in f(A1 v A2)

4. Let f:A->B and g:B->A be two functions that have inverses f' and g', 
respectively. Show, by direct inference, that function g.f has an inverse f'.g'.

Sol:
 T
=> {Definition of (f'.g')}
 (f'.g')(g.f(x)) =  f'(g'(g.f(x)))
=> {Definition of (g.f)}
 (f'.g')(g.f(x)) =  f'(g'(g(f(x))))
=> {g'(g(f(x))) = f(x)}
 (f'.g')(g.f(x)) =  f'(f(x))
=> {f'(f(x)) = x}
 (f'.g')(g.f(x)) =  x

5. Let R be the set of all real numbers, let f:R->R and g:R->R be two functions
where two functions g and g.f are defined as follows:
      	  g(x) = 2x+1 
	  g.f(x) = 2x+11
Use direct inference to derive the definition of function f.

Sol:
 T
=> {g(x) = 2x+1}
 g(x) = 2x+1
=> {Definition of g.f(x)}
 g.f(x) = g(f(x)) = 2*f(x)+1
=> {g.f(x) = 2x+11}
 2x+11 = 2f(x)+1
=> {Arithmetics}
 f(x) = x+5