Mohamed G. Gouda						     CS 313K
Fall 2012							     Exercise 4

1. Give a direct inference proof to prove the predicate: 
(All n, (n is odd) => (Exist k,l, n = k^2 - l^2))
where the domains of n, k, and l are the set of all positive integers.

Sol:
  n is odd
=> {definition of odd}
  n = 2m + 1, for some integer m
=> {add m^2 - m^2 to right side of "="}
  n = (m^2 + 2m + 1) - m^2, for some integer m
=> {arithmetics}
  n = (m+1)^2 - m^2, for some integer m
=> {rename m+1 and m}
  n = k^2 - l^2, for some integers k and l

2. Give a two-sided inference proof to prove the two-sided inference:
(All m, n, (m^2 = n^2) <=> (m=n or m=-n))
where the domains of m and n are the set of all real numbers.

Sol:
  m^2 = n^2
<=> {subtract n^2 from both sides of "="}
  m^2 - n^2 = 0
<=> {arithmetics}
  (m-n)(m+n) = 0
<=> {arithmetics}
  m-n = 0 or m+n = 0
<=> {arithmetics}
  m = n or m = -n
 
3. Give a two-sided inference proof to prove the two-sided inference:
(All n, n is even <=> (-1)^n = 1)
where the domain of n is the set of all non-negative integers.

Sol:
  n is even
<=> {definition of even}
  n = 2k, for some integer k, and (-1)^n = (-1)^(2k)
<=> {x^(2k) = (x^2)^k}
  n = 2k, for some integer k, and (-1)^n = ((-1)^2)^k
<=> {(-1)^2 = 1}
  n = 2k, for some integer k, and (-1)^n = 1^k
<=> {1^k = 1}
  n = 2k, for some integer k, and (-1)^n = 1

4. Give a direct inference proof to prove:
(All x, (x is rational and x != 0) => (1/x is rational))
where the domain of x is the set of all rational numbers.

Sol:
  (x is rational and x != 0)
=> {definition of rational}
  x = a/b, for some integers a and b where b != 0 and a and b share no factor 
  and x != 0
=> {arithmetics}
  x = a/b, for some integers a and b where b != 0 and a and b share no factor 
  and a != 0
=> {arithmetics}
  1/x = b/a for some integers a and b where a != 0 and a and b share no factor
=> {definition of rational}
  1/x is rational

5. Give an induction proof to prove that, for every non-negative integer n,
(All n, n >= 0, 7^0 + 7^1 + ... + 7^n = (7^(n+1) - 1)/6)

Sol:
Let P(n) be the predicate 
7^0 + 7^1 + ... + 7^n = (7^(n+1) - 1)/6

Base Case: n=0:
P(0) <=> 7^0 = (7^1 - 1)/6 <=> 1 = 1 <=> T

Induction Step: 
(All n, n >= 0, P(n) => P(n+1))

  P(n)
=> {definition of P(n)}
  7^0 + ... + 7^n = (7^(n+1) - 1)/6
=> {add 7^(n+1) to both sides of "="}
  7^0 + ... + 7^(n+1) = (7^(n+1) - 1)/6 + 7^(n+1)
=> {arithmetics}
  7^0 + ... + 7^(n+1) = (7^(n+1) - 1)/6 + 7^(n+1)
=> {arithmetics}
  7^0 + ... + 7^(n+1) = (7^(n+2) - 1)/6
=> {definition of P(n+1)}
  P(n+1)

6. Give an induction proof to prove the quantified predicate, 
(All n, n >= 1, 1/2 + 1/2^2 + ... + 1/2^n = (2^n - 1)/2^n)

Sol:
Let P(n) be the predicate 
1/2 + 1/2^2 + ... + 1/2^n = (2^n - 1)/2^n

Base Case: n=1:
P(1) <=> 1/2 = (2-1)/2 <=> 1/2 = 1/2 <=> T

Induction Step:
(All n, n >= 1, P(n) => P(n+1))

  P(n)
=> {definition of P(n)}
  1/2 + 1/2^2 + ... + 1/2^n = (2^n - 1)/2^n
=> {add 1/2^(n+1) to both sides of "="}
  1/2 + 1/2^2 + ... + 1/2^(n+1) = (2^n - 1)/2^n + 1/2^(n+1)
=> {arithmetic}
  1/2 + 1/2^2 + ... + 1/2^(n+1) = (2*(2^n - 1) + 1)/2^(n+1)
=> {arithmetic}
  1/2 + 1/2^2 + ... + 1/2^(n+1) = (2^(n+1) - 1)/2^(n+1)
=> {definition of P(n+1)}
  P(n+1)

7. Give an induction proof to prove the quantified predicate, 
(All n, n >= 1, (2*1 - 1) + (2*2 - 1) + ... + (2*n - 1) = n^2)

Sol:
Let P(n) be the predicate:
(2*1 - 1) + (2*2 - 1) + ... + (2*n - 1) = n^2

Base Case: n=1: 
P(1) <=> (2*1 - 1 = 1^2) <=> (1=1) <=> T

Induction Step:
(All n, n >= 1, P(n) => P(n+1))

  P(n)
=> {definition of P(n)}
  (2*1-1) + (2*2-1) + ... + (2*n-1) = n^2
=> {add 2*(n+1)-1} to both sides of "="}
  (2*1-1) + ... + (2*(n+1)-1) = n^2 + (2*(n+1)-1)
=> {arithmetics}
  (2*1-1) + ... + (2*(n+1)-1) = (n^2 + 2n + 1)
=> {arithmetics}
  (2*1-1) + ... + (2*(n+1)-1) = (n+1)^2
=> {definition of P(n+1)} 
  P(n+1)

8. Give an inducton proof to prove that, for every positive integer n,
(5^n - 1) is divisible by 4.

Sol:
Let P(n) be the predicate: (5^n - 1) is divisible by 4

Base Case: n = 1: 
P(1) <=> (5 - 1 is divisible by 4) <=> T

Induction Step: 
for every n > 0, P(n) => P(n+1)

  P(n)
=> {definitions of P(n) and of divisible by 4}
  5^n - 1 = 4k, for some integer k
=> {arithmetics}
  5^n = 4k+1, for some integer k
=> {multiply 5 to both sides of "="}
  5^(n+1) = 5*(4k+1), for some integer k
=> {arithmetics}
  (5^(n+1) - 1) = (5*(4k+1) - 1), for some integer k
=> {arithmetics}
  (5^(n+1) - 1) = 4*(5k+1), for some integer k
=> {definition of divisible by 4}
  (5^(n+1) - 1) is divisible by 4
=> {definition of P(n+1)}
  P(n+1)

9. Give an induction proof to prove that, for every non-negative integer n, 
(n^3 - n) is divisible by 3.

Sol:
Let P(n) be the predicate: (n^3 - n) is divisible by 3

Base Case: n=0:
P(0) <=> 0 is divisible by 3 <=> T

Induction Step:
for every n >= 0, P(n) => P(n+1)

  P(n)
=> {definitions of P(n) and divisible by 3}
  (n^3 - n) = 3k, for some integer k
=> {add (3n^2 + 3n) to both sides of "="}
  (n^3 - n) + (3n^2 + 3n) + 1 - 1 = 3k + (3n^2 + 3n)
=> {arithmetics}
  (n+1)^3 - (n+1) = 3k + 3n^2 + 3n
=> {(k + n^2 + n) is an integer}
  (n+1)^3 - (n+1) = 3m, for some integer m
=> {definitions of P(n+1) and of divisible by 3}
  P(n+1)

10. Give an induction proof to prove that, for every integer n > 10, 2^n > n^3.

Sol:
Let P(n) be the predicate 2^n > n^3.

Base Case: n=10: 
P(10) <=> (2^10 > 10^3) <=> (1024 > 1000) <=> T

Induction Step:
for every n >= 10, P(n) => P(n+1) 

  P(n)
=> 2^n > n^3
=> {multiply both sidex by 2}
  2^(n+1) > n^3*2
=> {arithmetics and n^3 >= 10*n^2 since n >= 10}
  2^(n+1) > (n^3 + 10*n^2)
=> {arithmetics and 7*n^2 >= 70*n since n >= 10}
  2^(n+1) > (n^3 + 3*n^2 + 70*n)
=> {arithmetics}
  2^(n+1) > (n^3 + 3*n^2 + 3*n + 67*n)
=> {arithmetics and 67*n >= 1 since n >= 10}
  2^(n+1) > (n^3 + 3*n^2 + 3*n + 1)
=> {arithmetics}
  2^(n+1) > (n+1)^3
=> {definition of P(n+1)}
  P(n+1)

11. Give an induction proof to prove that, for every positive integer n,
there are n(n-1)/2 item pairs in any set of n items.

Sol:
Let P(n) be the predicate: (there are n(n-1)/2 item pairs in any set of n items)

Base Case: n=1:
P(1) <=> (there are 0 item pairs in any set of 1 item) <=> T

Induction Step: 
for every n >= 1, P(n) => P(n+1)

  P(n)
=> {definition of P(n)}
  (there are n(n-1)/2 item pairs in any set of n items)
=> {add a new item t to any set of n items}
  (there are n(n-1)/2 item pairs that do not involve t in the set) +
  (there are n item pairs that involve t in the set)
=> {(n(n-1)/2 + n) = (n+1)n/2}
  (there are (n+1)n/2 item pairs in any set of n+1 items)
=> {definition of P(n+1)}
  P(n+1)

12. Define (n!) to be the multiplication (1*2*...*n). Give an induction proof 
to prove that, for every integer n >= 4, (2^n =< n!)

Sol:
Let P(n) be the predicate (2^n =< n!).

Base Case: n=4:
P(4) <=> (2^4 =< 4!) <=> (16 =< 24) <=> T

Induction Step: 
for every n >= 4, P(n) => P(n+1)

  P(n)
=> {definition of P(n)}
  (2^n =< n!)
=> {multiply 2 to both sides of "="}
  (2^(n+1) =< 2*n!)
=> {(n >= 4) => n+1 >= 2}
  (2^(n+1) =< (n+1)*n!)
=> {arithmetics}
  (2^(n+1) =< (n+1)!)
=> {definition of P(n+1)}
  P(n+1)

13. Verfiy that the following program specification holds:
	(x=X and y=Y)
	y := x+y
	(x=X and y=(X+Y))

Sol:
We need to prove the following predicate
	(All x,y, (x=X and y=Y) => (x=X and (x+y)=(X+Y)))

	(x=X and y=Y)
=> {Arithmetics}
	(x=X and y=Y and (x+y)=(X+Y))
=> {(P and Q and R) => (P and R)}
	(x=X and (x+y)=(X+Y))

14. Verify that the following program specification holds:
	(T)
	x:=5; y:=7
	(x=5 and y=7)

Sol:
We need to verify the following two program specifications:
	(T)
	x:=5
	(x=5)

	(x=5)
	y:=7
	(x=5 and y=7)

Thus we need to prove the following two predicates:
	(All x,y, (T) => (5=5))
	(All x,y, (x=5) => (x=5 and 7=7))

Proof of First Predicate:
	(T)
=> {5=5 is clearly T}
	(5=5)

Proof of Second Predicate:
	(x=5)
=> {P => (P and T)}
	(x=5 and T)
=> {7=7 is clearly T}
	(x=5 and 7+7)

15. Verify that the following program specification holds:
	(T)
	if x =< y then x := y else skip fi
	(x >= y)

Sol:
We need to verify the following two program specifications:
	(T and (x =< y))
	x := y
	(x >= y)

	(T and (x > y))
	skip
	(x >= y)

Thus we need to prove the following two predicates:
	(All x,y, (T and (x =< y)) => (y >= y))
	(All x,y, (T and (x > y)) => (x >= y))

Proof of First Predicate:
	(T and (x =< y))
=> {(P and Q) => P}
	(T)
=> {T => (y >= y)}
	(y >= y)

Proof of Second Predicate:
	(T and (x > y))
=> {(P and Q) => Q}
	(x > y)
=> {Arithmetics}
	(x >= y)

16. Verfiy that the following program specification holds:
	(x > 10)
	skip
	(x >= 10)

Sol:
We need to prove the following predicate
	(All x,y, (x > 10) => (x >= 10))

	(x >10)
=> {P => (P or Q)}
	(x >10 or x =10)
=> {arithmetics}
	(x >= 10)

17. Verfiy that the following program specification holds:
	(x >= 0)
	x := x+1
	(x > 0)

Sol:
We need to prove the following predicate
	(All x,y, (x >= 0) => ((x+1) > 0))

	x >= 0
=> {adding 1 to both sides of >= maintains >=}
	(x+1) >= 1
=> {arithmetics}
	(x+1) > 0

18. Verfiy that the following program specification holds:
	(y = 10)
	x := y
	(y > 5)

Sol:
We need to prove the following predicate
	(All x,y, (y=10) => (y>5))

	y = 10
=> {arithmetics}
	y > 5

19. Verfiy that the following program specification holds:
	(T)
	x := x+y
	(x >= y)

Sol:
We need to prove the following predicate
	(All x,y, (T) => ((x+y) >= y))

	T
=> {(x+y) >= y is clearly true}
	(x+y) >= y

20. Verfiy that the following program specification holds:
	(x=5 and y=2)
	y := x+1
	(y > 5)

Sol:
We need to prove the following predicate
	(All x,y, (x=5 and y=2) => ((x+1) > 5)

	x=5 and y=2
=> {(P and Q) => P}
	x=5
=> {arithmetics}
	(x+1) > 5

21. Verify that the following program specification holds:
	(y = Y)
	x := (y+1); x := (y+2) 
	(x = (Y+2))

Sol:
We need to verify the following two program specifications:
	(y = Y)
	x := y+1
	(y = Y)

	(y = Y)
	x := y+2
	(x = (Y+2))

Thus we need to prove the following two predicates:
	(All x,y, (y = Y) => (y = Y))
	(All x,y, (y = Y) => ((y+2) = (Y+2))

Proof of First Predicate:
	(y = Y)
=> {P => P}
	(y = Y)

Proof of Second Predicate:
	(y = Y))
=> {adding 2 to both sides of = maintains =}
	((y+2) = (Y+2))

22. Verify that the following program specification holds:
	(T)
	if x =! y then y := x else skip fi
	(x = y)

Sol:
We need to verify the following two program specifications:
	(T and (x =! y))
	y := x
	(x = y)

	(T and not(x =! y))
	skip
	(x = y)

Thus we need to prove the following two predicates:
	(All x,y, (T and (x =! y)) => (x = x))
	(All x,y, (T and not(x =! y)) => (x = y))

Proof of First Predicate:
	(T and (x =! y))
=> {(P and Q) => P}
	(T)
=> {(x=x) is clearly true}
	(x = x)

Proof of Second Predicate:
	(T and not(x =! y))
=> {(arithmetics}
	(T and (x = y))
=> {(P and Q) => Q}
	(x = y)