Move constructor on derived object

rval is not a Rvalue. It is an Lvalue inside the body of the move constructor. That's why we have to explicitly invoke std::move.

Refer this. The important note is

Note above that the argument x is treated as an lvalue internal to the move functions, even though it is declared as an rvalue reference parameter. That's why it is necessary to say move(x) instead of just x when passing down to the base class. This is a key safety feature of move semantics designed to prevent accidently moving twice from some named variable. All moves occur only from rvalues, or with an explicit cast to rvalue such as using std::move. If you have a name for the variable, it is an lvalue.

Named R-value references are treated as L-value.

So we need std::move to convert it to R-Value.