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eml(1,eml(x,1)) = e + x

and

eml(eml(1,x),1) = e^e * x



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




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