Okay, I’m tired. Not quite inverse but per the title , must be a way.
I was mistaken above in the first identity, it is
eml(1,eml(x,1)) = e - x
Which then if you iterate gives x (ie is inverse of itself).
eml(1,eml(eml(1,eml(x,1)),1)) = x
alt Hacker News
I was mistaken above in the first identity, it is
eml(1,eml(x,1)) = e - x
Which then if you iterate gives x (ie is inverse of itself).
eml(1,eml(eml(1,eml(x,1)),1)) = x