Solving a Smullyan style knight and knave problem using natural deduction. How to shorten this proof? [ EDITED, there were 2 useless lines in the deduction below]
The Smullyan style problem : 
John: If (and only if) Bill is a knave, then I am a knave.
Bill: We are of different kinds.
With the understanding that knights only tell true statements and knaves only tell false statements, identify John and Bill's types.
(It therefore follows that if $J$ is the proposition that John is a knight, and John says a logical statement $X$, that $J\leftrightarrow X$ is a tautology.)
My attempt uses natural deduction; but the reasoning is rather long. 
Can you think of a shorter natural deduction style proof? 
What would be the quickest way to solve the problem according to you? 
Below , "J" means " John is a knight" and "B" means " Bill is a knight". 
Note: this solution makes Bill's statement true and John's statement false; which is coherent with their being a knight and a knave respectively. 
Indeed if we have ( B & ~J) ( that is our alledged solution) John's statement is false, since this statement is equivalent to ( B <-> J) , which, in turn, is equivalent to ~ (B&~J) & ~ ( J&~B). 


I understand that this is not a natural deduction proof, but in response to benrg's comment above, I read somewhere else that you can use algebra in the Galois field of two elements GF(2) to solve things like these, and in that case if you take Z="A is a knave" then you would formalize "A asserts p" as Z + p = 1, since either Z or p are true, but not both . I don't know the details of how to scale this for difficult problems but for the OP's problem, a solution is pretty short:

*

*Take: A = "A is a knight", B = "B is a knight", Z = "A is a knave" and Y = "B is a knave"

*By definition: A+Z = B+Y = 1

*Inclusive p or q is "p + q - pq = 1"

*The claims give us the equations: Z + AB + YZ - ABYZ = 1 and Y + AY + BZ - ABYZ = 1

*ABYZ = 0 so Z + AB + YZ = Z(1 + Y) + AB = 1 and Y + AY + BZ = Y(1 + A) + BZ = 1

*Using (2) (1+Y) = B and (1+A)=Z, substituting in (5) we end up with: B(Z+A)=1 and Z(Y+B)=1

*Then B=1 and Z=1, that is, A is a knave and B is a knight.

