In the expression left() = right(), why is right() sequenced first?

In the proposal P0145 that introduced this evaluation order, the authors gave the following example:

#include <map>
int main() {
    std::map<int, int> m;
    m[0] = m.size();
}

In this situation, left-to-right evaluation would give 1, whereas right-to-left evaluation would give 0. Having the result be 0, thanks to right-to-left evaluation, corresponds more closely to our intuition that the value that should be assigned is the one that existed immediately before the assignment expression was evaluated.

In addition to the unintuitive result when doing what Brian showed:

#include <map>
int main() {
    std::map<int, int> m;
    m[0] = m.size(); // before C++17 m[0] could be 0 or 1 - it was implementation defined
}

If we take a the same map but do:

#include <map>
int main() {
    std::map<int, int> m;
    m[0] = Right(); // Right() may throw
}

If Right() throws:

Before C++17 you could get a default constructed element in m[0] (left to right) or m[0] wouldn't be created at all (right to left). In C++17 m[0] will not get created at all.