Understand std::move & std::unique_ptr in C++

Programming
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1

I am trying to understand the differences in the behavior of std::move and the transfer of ownership with std::unique_ptr when used with functions F1 and F2. The code is given below:

void F1(std::unique_ptr<Dog>&& uPtr)
{
    std::cout << "F1 \n";
}

void test1()
{
    std::unique_ptr<Dog> pD(new Dog("Gunner"));
    F1(std::move(pD));
    if (pD == nullptr)
    {
        std::cout << "Null\n";
    }
    std::cout << "Test \n";
}

void F2(std::unique_ptr<Dog> uPtr)
{
    std::cout << "F2 \n";
}

void test2()
{
    std::unique_ptr<Dog> pD(new Dog("Smokey"));
    F2(std::move(pD));
    if (pD == nullptr)
    {
        std::cout << "Null\n";
    }
    std::cout << "Test \n";
}

When std::move(pD) is used to pass pD to F1 in test1, pD is not null after the call. However, when std::move(pD) is used to pass pD to F2 in test2, pD is null after the call. Why is this behavior different?