How does one actually add with this?
Well, once you've derived unary exp and ln you can get subtraction, which then gets you unary negation and you have addition.
Don't know adding, but multiplication has diagram on the last page of the PDF.
xy = eml(eml(1, eml(eml(eml(eml(1, eml(eml(1, eml(1, x)), 1)), eml(1, eml(eml(1, eml(y, 1)), 1))), 1), 1)), 1)
From Table 4, I think addition is slightly more complicated?
It's basically using the "-" embedded in the definition of the eml operator.
Table 4 shows the "size" of the operators when fully expanded to "eml" applications, which is quite large for +, -, ×, and /.
Here's one approach which agrees with the minimum sizes they present:
After you have ln and exp, you can invert their applications in the eml function Using a subtraction-of-subtraction to get addition leads to the cost of "27" in Table 4; I'm not sure what formula leads to 19 but I'm guessing it avoids the expensive construction of 0 by using something simpler that cancels: