[...]

I'm confused about what you are asking me to prove. If I understand correctly, my only evidence is that the standard currently makes those assertions.

[...] The actual problem nests, kinda like swiss cheese. Because, you can have this...

```c++

struct Foo

{

    Bar bar;

    Baz baz;

};

static_assert(std::is_implicit_lifetime_type_v<Bar>);

static_assert(std::is_implicit_lifetime_type_v<Baz>);

static_assert(not std::is_implicit_lifetime_type_v<Foo>);

```

The only way that Foo will be an implicit lifetime type under the current definition is if both Bar and Baz have at least one trivial constructor IN COMMON.

A class S is an implicit-lifetime class if
- it is an aggregate whose destructor is not user-provided or

- it has at least one trivial eligible constructor and a trivial, non-deleted destructor.

Scalar types, implicit-lifetime class types, array types, and cv-qualified versions of these types are collectively called implicit-lifetime types.