grammar - Can 「食べも飲みもしない」 be rewritten as 「食べなくも飲まな くもある」?

I'd like to see if I understand a couple grammar points correctly. I'd like to rewrite this sentence:

 1. 食べも飲みもしない

As either one of these:

 2a. 食べなくも　　飲まなくもある
 2b. 食べなくもあり飲まなくもある

I think both 2a and 2b are grammatical with the same meaning as 1. But I'm not sure I've done this correctly, and I'm particularly unsure about whether the あり is necessary in the first half.

---

The following is my reasoning. (I've tried to express it as clearly as I can, but it's probably too long and confusing--my apologies!)

I'm working off Hiroshi Aoyagi's idea that -i is a suppletive form of -ku ar-(r)u, so the end of the sentence looks like this:

　　*飲みも　　 しなくある　　 　→　飲みも　　 しない
   *nom(i)-mo si-na-ku ar-u　→　nom(i)-mo si-nai

Here, the stem si appears because the suffix -(a)na-ku needs to have something to attach to, and it can't attach directly to the verb stem nom- because -mo is in the way.

I'm moving -(a)na-ku before -mo and attaching it directly to both verb stems, so si is no longer necessary and it disappears. But the tense morpheme -(r)u still needs to appear at the end of the sentence, so I have to insert the dummy verb ar- for it to attach to:

　　飲まなくも　　　　ある
　　nom-(a)na-ku-mo ar-u

Since -mo appears between -ku and ar-(r)u, it can't take the usual form of -i.

Since the tense morpheme -(r)u only has to appear once at the end of the sentence, I don't think I need to add the dummy verb ar- to the first half. But if I do, I think it ends up in "infinitive" form (あり):

　　食べなくも　　　   　　 飲まなくも　　　　ある
　　tabe-na-ku-mo        nom-(a)na-ku-mo ar-u


　　食べなくも　　　あり　　 飲まなくも　　　　ある
　　tabe-na-ku-mo ar-(i) nom-(a)na-ku-mo ar-u

So based on this, I suspect that both 2a and 2b are equivalent to 1. Am I correct or mistaken?