4 ms·

x+y = ln(exp(x) * exp(y)) exp(a) = eml(a, 1) ln(a)=eml(1,eml(eml(1,a),1)) Plugging those in is an excercise to the reader

• jcgrillo 6mo ago
  might need to turn the paper sideways
• freehorse 6mo ago
  You use multiplication in the first line, which you have not expressed through eml yet. Because of how exp and log turn addition into multiplication and vice versa, once you have the one, you get the other easily.