Mohamed G. Gouda						     CS 313K
Fall 2012							     Homework 1

1. Give a direct inference proof to prove 
  m+n and n+p are even => m+p is even
where the domains of m, n, and p are the set of all integers.

Sol:
  m+n and n+p are even
<=> {Definition of even}
  m+n = 2k and n+p = 2l, for some integers k and l
=> {Add m+n to n+p}
  m+2n+p = 2 (k+l), for some integers k and l
<=> {Arithmetics}
  m+p = 2(k+l-n), for some integers k and l
=> {k+l-n is an integer}
  m+p = 2q, for some integer q
<=> {Definition of even}
  m+p is even

2. Give a by-contradiction proof to prove 
  All m, n (m^2 - n^2 != 2)
where the domains of m and n are the set of all positive integers.

Sol:
  m^2 - n^2 = 2
<=> {Arithmetics}
  (m-n)(m+n) = 2
<=> {m,n,(m-n), and (m+n) are all positive integers}
  m-n = 1 and m+n = 2  
=> {Solving for m}
  m = 3/2
=> {Contradicting that m is a positive integer}
  F

3. Give an indirect inference proof to prove
  m*n is even => m is even or n is even
where the domains of m and n are the set of all integers.

Sol:
  not (m is even) and not (n is even)
<=> {every integer is either even or odd}
  m is odd and n is odd
<=> {Definition of odd}
  m = 2k+1 and n = 2l+1, for some integers k and l
=> {Multiply m*n}
  m*n = (2k+1)(2l+1), for some integers k and l
<=> {Arithmetics}
  m*n = 4kl + 2k + 2l + 1 = 2 (2kl + k + l) + 1, for some integers k and l
=> {2kl + k + l is an integer}
  m*n = 2p + 1, for some integer p
<=> {Definition of odd}
  m*n is odd

4. Give two inference proofs that together prove the following two-sided 
inference:
  n is even <=> (7*n + 4) is even
where the domain of n is the set of all positive integers.
(Hint: prove by direct inference: n is even => (7*n + 4) is even.
Then, prove by indirect inference: n is odd => (7*n + 4) is odd.)

Sol:
(Proof of =>)

  n is even
<=> {Definition of even}
  n = 2k, for some integer k
=> {Compute 7*n + 4}
  7*n + 4 = 14*k + 4, for some integer k
<=> {Arithmetics}
  7*n + 4 = 2*(7k + 2), for some integer k
=> {7k + 2 is an integer}
  7*n + 4 = 2l, for some integer l
<=> {Definition of even}
  7*n + 4 is even

(Proof of <=)

  n is odd
<=> {Definition of odd}
  n = 2k+1, for some integer k
=> {Compute 7*n + 4}
  7*n + 4 = 14*k + 11, for some integer k
<=> {Arithmetics}
  7*n + 4 = 2*(7k + 5) + 1, for some integer k
=> {7k + 5 is an integer}
  7*n + 4 = 2l + 1, for some integer l 
<=> {Definition of odd}
  7*n + 4 is odd

5. Give an induction proof to prove that for every integer n, where n >= 2, 
   not (f.1(x) or f.2(x) or ... or f.n(x)) = 
   not f.1(x) and not f.2(x) and ... and not f.n(x)   

Sol:
  Let P(n) be the predicate:
   not (f.1(x) or f.2(x) or ... or f.n(x)) = 
   not f.1(x) and not f.2(x) and ... and not f.n(x)   
  
Base Case (n = 2):
  P(2) = T by De-Morgan's law
  
Induction Step:
  P(n) => P(n+1)    

  P(n)

<=> {Definition of P(n)}
  not (f.1(x) or f.2(x) or ... or f.n(x)) = 
  not f.1(x) and not f.2(x) and ... and not f.n(x) 

<=> {Add "and not f.n+1(x)" to both sides of "="}
  not (f.1(x) or f.2(x) or ... or f.n(x)) and not f.n+1(x) = 
  not f.1(x) and not f.2(x) and ... and not f.n(x) and not f.n+1(x)
  
<=> {Apply De-Morgan's law to the left hand side of "="}
  not (f.1(x) or f.2(x) or ... or f.n(x) or f.n+1(x)) =
  not f.1(x) and not f.2(x) and ... and not f.n(x) and not f.n+1(x)

<=> {Definition of P(n+1)}
  P(n+1)

6. Give an induction proof to prove that for every positive integer n,
        1^2 + 2^2 + ... + n^2 = 1/6*n*(n+1)*(2n+1)

Sol:
  Let P(n) be the predicate:
        1^2 + 2^2 + ... + n^2 = 1/6*n*(n+1)*(2n+1)

Base Case(n=1):
  P(1) <=> (1^2 = 1/6*1*(1+1)*(2*1+1)) <=> (1 = 1/6*2*3) <=> (1 = 1) <=> T

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

  P(n)

<=> {Definition of P(n)} 
  (1^2 + 2^2 + ... + n^2 = 1/6*n*(n+1)*(2n+1))

<=> {Add (n+1)^2 to both sides of "="}
  (1^2 + 2^2 + ... + n^2 + (n+1)^2 = 1/6*n*(n+1)*(2n+1) + (n+1)^2)

<=> {Arithmetics to the right hand side of "="}
  (1^2 + 2^2 + ... + n^2 + (n+1)^2 = 1/6*(n+1)*(n*(2n+1) + 6(n+1))

<=> {Arithmetics to the right hand side of "="}
  (1^2 + 2^2 + ... + n^2 + (n+1)^2 = 1/6*(n+1)*(2n^2 + 7n + 6))

<=> {Arithmetics to the right hand side of "="}
  (1^2 + 2^2 + ... + n^2 + (n+1)^2 = 1/6*(n+1)*(n + 2)*(2n + 3))

<=> {Definition of P(n+1)}
  P(n+1)