why the rvalue is deduced to be lvalue?

Question

The code is attach as follow. i tried to implement a method to check the expression or varible's value type. but it seems failed to check rvalue.

#include <iostream>
#include <type_traits>
#include <vector>
#include <string>

template <typename T>
void checkLValueRValue(T&& value) {
    using Type = decltype(std::forward<T>(value));
    if (std::is_lvalue_reference<Type>::value) {
        std::cout << "It's an lvalue reference." << std::endl;
    } else if (std::is_rvalue_reference<Type>::value) {
        std::cout << "It's an rvalue reference." << std::endl;
    } else {
        std::cout << "It's not an lvalue or rvalue reference." << std::endl;
    }
}

class A {

};


int main() {
    int x = 42;
    int y = 43;
    int& lvalue_ref = x;
    int&& rvalue_ref = std::move(x);
    std::cout << std::is_rvalue_reference_v<A&&> << std::endl;

    checkLValueRValue(x);          // lvalue reference
    checkLValueRValue(lvalue_ref); // lvalue reference
    checkLValueRValue(std::move(x));// rvalue reference
    *checkLValueRValue(rvalue_ref); // it should be rvalue reference but not!*

    return 0;
}

the last checkLValueRvalue outputs It's an lvalue reference ? what happened ?

i am new to cpp, so i don't get it.