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Mohamed G. Gouda						     CS 311
Fall 2014							     Midterm 1
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1. (2 points)
Write the Base Case proof that is part of the induction proof for the predicate
	(All x, x >= 6, 1+2+..+x = x(x+1)/2)
where the domain of x is the set of all integers

2. (3 points)
Use Equivalence Laws to prove that the following two formulas are equivalent: 
	T --> f(x)
	f(x)

3. (5 points)
Use Equivalence Laws to prove that the following two formulas are equivalent: 
	(f(x) --> g(x)) and (g(x) --> f(x))
	(f(x) = g(x)) 

4 and 5. (10 points)
Let	Dx be the set of all non-negative integers
	Dm be the set of all integers larger than 1
	(x mod m) be defined as (the remainder of dividing x by m)

4. (5 points)
Use direct inference to prove the following predicate
	(All x, (0 =< x and x < m) ==> (x = (x mod m)))

5. (5 points)
Use guessing to prove the following predicate
	(Exist x, (x mod m) > ((x+1) mod m)

Solutions:

1.
Let P(x) be the predicate (1+2+..+x = x(x+1)/2)

Base Case: x=6:
	P(6)
<=> {definition of P(6)}
	1+2+..+6 = 6(7)/2
<=> {arithmetics}
	21 = 3(7)
<=> {arthmetics}
	T

2.
	T --> f(x)
==> {definition of -->}
	not(T) or f(x)
==> {truth table of not}
	F or f(x)
==> {elementary}
	f(x)

3.
	(f(x) --> g(x)) and (g(x) --> f(x))
==> {definiton of -->}
	(not(f(x)) or g(x)) and (not(g(x)) or f(x))
==> {distribution of and over or}
	((not(f(x)) or g(x)) and not(g(x))) or
	((not(f(x)) or g(x)) and f(x))
==> {distribution of and over or}
	((not(f(x)) and not(g(x))) or (g(x) and not(g(x))) or
	((not(f(x)) and f(x)) or (g(x) and f(x))
==> {idempotence}
	((not(f(x)) and not(g(x))) or F) or
	(F or (g(x) and f(x)))
==> {elementary}
	(not(f(x)) and not(g(x))) or
	(g(x) and f(x))
==> {definition of =}
	f(x) = g(x)

4.
	(0 =< x and x < m) 
==> {definition of remainder}
	(x = the remainder of dividing x by m)
==> {definition of (x mod m)}
	(x = x mod m)

5.
For x = m-1, (x mod m) = m-1 > 0 = ((x+1) mod m)
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------------------------------------------------------------------------------
Mohamed G. Gouda						     CS 311
Fall 2014							     Midterm 1
------------------------------------------------------------------------------

1. (2 points)
Write the Base Case proof that is part of the induction proof for the predicate
	(All x, x >= 3, (2^0 + 2^1+..+ 2^x) = (2^(x+1) - 1))
where the domain of x is the set of all integers

2. (3 points)
Use Equivalence Laws to prove that the following two formulas are equivalent: 
	(f(x) and  not(g(x))) --> F
	(f(x) --> g(x)

3. (5 points)
Use Equivalence Laws to prove that the following two formulas are equivalent: 
	f(x) --> g(x)
	not(g(x) --> not(f(x))

4 and 5. (10 points)
Let	Dx be the set of all non-negative integers
	Dm be the set of all integers larger than 1
	(x mod m) be defined as (the remainder of dividing x by m)

4. (5 points)
Use direct inference to prove the following predicate
	(All x, (0 =< x and x < m) ==> (x = (x mod m)))

5. (5 points)
Use direct inference to prove the following predicate
	(All x, T ==> ((x mod m) = ((x mod m) mod m)))

Solutions:

1.
Let P(x) be the predicate (2^0 + 2^1 +..+ 2^x = 2^(x+1) - 1)

Base Case: x=3:
	P(3)
<=> {definition of P(3)}
	(1+2+4+8) = (2^4 - 1)
<=> {arithmetics}
	15 = 16 - 1
<=> {arthmetics}
	T
	

2.
	(f(x) and not(g(x))) --> F
==> {definition of -->}
	not(f(x) and not(g(x))) or F
==> {elementary}
	not(f(x) and not(g(x)))
==> {De Morgan's}
	not(f(x)) or not(not(g(x)))
==> {idempotence of not}
	not(f(x)) or g(x)
==> {definition of -->}
	f(x) --> g(x)

3.
	f(x) --> g(x)
==> {definition of -->}
	not(f(x)) or g(x)
==> {symmetry of or}
	g(x) or not(f(x))
==> {idempotence of not
	not(not(g(x))) or not(f(x))
==> {definition of -->}
	not(g(x)) --> not(f(x))

4.
	(0 =< x and x < m) 
==> {definition of remainder}
	(x = the remainder of dividing x by m)
==> {definition of (x mod m)}
	(x = x mod m)

5.
	T
==> {definition of (x mod m)}
	(x mod m) = (remainder of dividing x by m)
==> {definition of remainder}
	(0 =< (x mod m) and (x mod m) < m)
==> {predicate in Problem 4 above}
	(x mod m) = ((x mod m) mod m)
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