# Given x ⊕ x =5,x ⊕ (y ⊕ z) = (x ⊕ y)+z-5 what is 2009 ⊕ 1949

There is an operation "⊕" that satisfies

1. $$x ⊕ x =5$$
2. $$x ⊕ (y ⊕ z) = (x ⊕ y)+z-5$$

what is 2009 ⊕ 1949?
I think it is feasible to solve with Mathematica, but my code doesn't give a useful result, am i missing something?

Clear["*"];
f[2009, 1949] //. {f[x_, y_] :> f[x, f[y, z]] - z + 5, f[x_, x_] :> 5,
f[x_, f[y_, y_]] :> y + 5}

• With CirclePlus[2009, 1949] you can get the exact writing that your question has. CirclePlus is an operator with no built-in meaning. – Roman Oct 22 at 12:10
• f[2009, 1949]==65 – Ulrich Neumann Oct 22 at 14:27

This is the kind of thing that you can only find out by exploring substitutions. Doing a bit of trial in the 2nd condition I found a couple of relations that can solve the problem by substituting z for y, and then substituting both y and z for x.

x ⊕(y⊕z) == (x⊕y) + z - 5 /.
z -> y //. CirclePlus[x_, x_] :> 5
(* x⊕5 == -5 + y + x⊕y *)
x ⊕(y⊕z) == (x⊕y) + z - 5 /. {y ->
x, z -> x} //. CirclePlus[x_, x_] :> 5
(* x⊕5 == x *)


From the first relation we can easily get that x⊕y==5 - y + x⊕5; from the second sit is clear that this expression can be further simplified into x⊕y==5 - y + x.

Finally using this as a single rule applied to 2009 and 1949

2009⊕1949 /. {x_⊕y_ :> x + 5 - y}
(*65*)


This is the result that Ulrich Neumann has mentioned in the comments, and ⊕ is the operator that Roman has suggested in the comments as well.

There are several possible methods to try. Here is one that I came up with in a few minutes without knowing the answer. It depends on using a range of substitutions which may give the answer and it is possible that it may be improved.

First, I decided to use f instead of CirclePlus and V instead of $$5$$ for clarity and no loss of generality. The two properties of the operation are given to us. I used the the first as a replacement rule because it is so obvious and simple. I used the second as a template and substituted the three variables with a range of substitutions. My code:

Select[ Flatten[ Table[
f[i, f[j, k] == f[i, j] + k - V /. f[x_, x_] :> V,
{i, {x}}, {j, {x,y}}, {k, {x,y,z,V}}], 2],
FreeQ[#, f[_, f[_, _]]]&]


returns the result

{f[x, V] == x, f[x, V] == -V + y + f[x, y]}
`

and combining the two equalities it is immediate that $$f(x,y) = x - y + V.$$.