I think the description is clear that N=0 means the result of a binary fold is E.

For the first point, I think I agree the presentation could be improved by first introducing E_i (in general) and then explaining where those appear.

result

For the second point, I don't think it's clear. In paragraph 10, the rule only states how the result would be for a unary fold-expression when the element of the pack expansion parameter is empty, It does not talk about binary fold-expression in this case.

Although there's a rule possibly relate to the case, which inhabits at the end of paragraph 8, that is:

Such an instantiation does not alter the syntactic interpretation of the enclosing construct, even in cases where omitting the list entirely would otherwise be ill-formed or would result in an ambiguity in the grammar.

However, at this time, I think the rule prefers to describe the case for "The instantiation of a pack expansion that is neither a sizeof... expression nor a fold-expression..." when N is zero.

Consider this example

#include <iostream>
template<class...T>
void fun(T...args){
    std::initializer_list<bool> c = {true&&args...};
    std::cout<< c.size()<<std::endl;  // its result is 0
    std::initializer_list<bool> c2 = {(true&&...&&args)};
    std::cout<< c2.size()<<std::endl; // its result is 1
}

int main() {
    fun();
}

The size of c is zero rather than one element which has the value true. It's quite different with binary fold-expression. So, I think it's necessary to state the different results between a binary fold-expression and a normal expansion when their N is zero.

I'm saying that the syntactic expansion described for binary fold-expressions in p10 does work for N=0, and thus N=0 does not need a special case in the prose wording.

I'm saying that the syntactic expansion described for binary fold-expressions in p10 does work for N=0, and thus N=0 does not need a special case in the prose wording.

Maybe I see your meaning. Do you mean _op and E_i are pair-wise?

Sort-of. Look at the expansion of a unary fold:

((E 1 op E 2 ) op · · · ) op E N

What's the result for N=1? It's E_1 (there is no E_2 at all).

Now look at binary fold:

(((E op E 1 ) op E 2 ) op · · · ) op E N

What's the result for N=0? With analogy to the foregoing, it's obviously E (there is no E_i at all).

Sort-of. Look at the expansion of a unary fold:

((E 1 op E 2 ) op · · · ) op E N

What's the result for N=1? It's E_1 (there is no E_2 at all).

Now look at binary fold:

(((E op E 1 ) op E 2 ) op · · · ) op E N

What's the result for N=0? With analogy to the foregoing, it's obviously E (there is no E_i at all).

Thanks. That's my understanding of what you are saying. That is, from right to left the _op and its subsequent expression are viewed as an instantiated component.