From the Putnam Competition (I believe):

*Suppose # is a binary operation on a set S such that the following properties hold:*

- For all a, b, c in S, (a#b)#c = a#(b#c);
- For all a, b in S, if a#b = b#a then a=b.

Prove that, for all x, y, z in S, we have x#y#z = x#z.

First, note that by rule 1, x#(x#x) = (x#x)#x. So by rule 2, we have x#x = x for all x in S.

It follows that x#(x#y#x) = x#x#y#x = x#y#x = x#y#x#x = (x#y#x)#x, so that by rule 2, we have x#y#x = x for all x, y in S.

Finally, we now have (x#z)#(x#y#z) = x#z#x#y#z = x#y#z = x#y#z#x#z = (x#y#z)#(x#z), so that again by rule 2 we have x#y#z = x#z, and the result is proven.

What I especially like about this problem is that with very few conditions, we end up proving something very strong – that a lot of information in a string of elements is redundant. a#b#c#d#e#f#g is exactly the same element as a#g: the middle five terms add nothing!

As an aside, see if you can find an example of such an operation…

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