Understanding lvalue/rvalue expression vs object type

I've read some of the prior top answers as well as Stroustrup's "The C++ Programming Language" and "Effective Modern C++" but I'm having trouble really understanding the distinction between the lvalue/rvalue aspect of an expression vs its type. In the introduction to "Effective Modern C++" it says:

A useful heuristic to determine whether an expression is an lvalue is to ask if you can take its address. If you can, it typically is. If you can't, it's usually an rvalue. A nice feature of this heuristic is that it helps you remember that the type of an expression is independent of whether the expression is an lvalue or rvalue ... It's especially important to remember this when dealing with a parameter of rvalue reference type, because the parameter itself is an lvalue.

I'm not understanding something because I don't understand why if you have an rvalue reference type parameter you need to actually cast it to an rvalue via std::move() to make it eligible to be moved. Even if the parameter (all parameters) is an lvalue the compiler knows its type is an rvalue reference so why the need to tell the compiler that it can be moved? It seems redundant but I guess I am not understanding the distinction between the type of an expression vs its lvalue/rvalue nature (not sure of the right terminology).