User blog comment:Nayuta Ito/faketest/e0/@comment-30754445-20180804174000/@comment-35392788-20180804230945

First of all, you don't need to be act like a douche over it.

Then, yes, I meant U_1. That being said, it's not equal to what you said.

U_1(W_2) = e(W+1)

U_1(W_2+1) = e(W+1)*w

so U(e(W+1)w) is U(U_1(W_2+1))

It is not too hard to show that U(W_2+a) = BHO*w^a for countable a, which is equivalent to U(e(W+1)*w^a)

Now we have U(W_2+U_1(W_2+1)) = U(e(W+1)*w^(e(W+1)*w)).

We derive U(e(W+1)*(w^e(W+1))^w) = U(e(W+1)*e(W+1)^w)

Finally, U(e(W+1)^(1+w)) = U(e(W+1)^w)