I'm new to lambda calculus and am having trouble understand how a shorthand acts when reducing.
Given: $((\lambda xy.x)(\lambda y.y))y$
What I have:
since $\lambda xy.x = \lambda x.\lambda y.x$
$((\lambda x.\lambda y.x)(\lambda y.y))y$
Now from here, I'm not sure how to continue with the reduction. I believe the whole thing is supposed to simplify to $xy$ but I'm not sure how to get there.