the lifetime of the variant members of union class objects operated by memcpy

Consider this example

union U{
   int i;
   char c;
};
U u1 = {.i = 1};  // #1
U u2 = {.c = 0};  // #2
std::memcpy(&u2, &u1, sizeof(U));  // #3

At #1, the variant member U::i of the object u1 is active, while at #2, the variant member U::c of the object u2 is active. Since [cstring.syn] p3 says

The functions memcpy and memmove are signal-safe. Both functions implicitly create objects ([intro.object]) in the destination region of storage immediately prior to copying the sequence of characters to the destination.

Thus, before the copying starts, what can be sure is that an object of type U is implicitly created in the specified storage. In addition, the lifetime of either U::i or U::c can start due to the "implicitly create objects" but not both. Which is regulated by [class.union.general] p2

At most one of the non-static data members of an object of union type can be active at any time, that is, the value of at most one of the non-static data members can be stored in a union at any time.

Since, immediately prior to copying, we cannot determine which variant member has begun its lifetime in the newly created union object that occupies the storage of u2. Also, after the copying, we also cannot determine which one is active. Is it U::i because the corresponding variant is active in the source? Or, Is it U::c since it's the one selected by the "implicitly creating objects" operation.

Start with the issue's union example and read the cited [cstring.syn], [intro.object], and [class.union.general] paragraphs. Determine how the active variant member's lifetime should be interpreted before and after memcpy, and clarify the standard wording so the resulting state is unambiguous.