This should be very easy on paper (those who play such games can do it most likely in their head). But how will it go in real time on-line? It feels to me awkward but let me try for awhile.

First try

Axioms

Let’s consider system  (X -),  where  –  is a binary operation in  X,  which satisfies the following axioms:

1. x – (y – z)  =  z – (y – x)
2. x-x = y-y
3. x – (y-y)  =  x

for every  x y z ∈ X.

Consequences

From axioms  1. and 3.  (let z:=y)  we get:

y – (y – x) = x

Let’s define:

• 0  :=  x-x     (the choice of  x  is irrelevant)
• Neg(x)   :=  0 – x
• x+y  :=  x – Neg(y)

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Wen can rewrite axioms 3 as:

x – 0 = x

THEOREM 0

Neg(Neg(x)) = x

PROOF

Neg(Neg(x))  =  (z-z) – ((y-y) – x)  =  x -((y-y) – (z-z)  =  x

END of PROOF

THEOREM 1

x+y = y+x

PROOF

x+y  =  x – (0 – y)  =  y – (0 – x)  =  y+x

END of PROOF

THEOREM 2

x+0 = x

PROOF

x+0  =  x – (0 – 0)  =  x – 0  =  x

END of PROOF

THEOREM 3

(x+y)+z  =  x+(y+z)

PROOF

(x+y)+z  =  (x – (0 – y)) – (0 – z)  =  z – (0 – (x – (0 – y))

z – ((0 – y) – (x – 0))  =  z – ((0 – y) – x)

By the same token (swap  x  and  z):

(z+y)+x  =  x – ((0 – y) – z)

But

x – ((0 – y) – z)  =  z – ((0 – y) – x)

hence

((x+y)+z  =  ((z+y)+x

and we already showed that operation + is commutative–therefore

((x+y)+z  =  (x+(y+z)

END of PROOF

THEOREM 4

x-y  =  x + Neg(y)

PROOF

x + Neg(y)  =  x – Neg(Neg(y))  =  x-y

END of PROOF

Ooooph

Oooph, I got lucky :-). It was hard for me to type, copy/paste, and manage html, and at the same time to think. Fortunately, I’ve selected the axioms well, and then there was very little left to think about, lucky me.

Now the road to integland is wide open–hurray!