The type of a pointer to a class method using deducting this matches the type of a pointer to a free function:

#include 

struct TA
{
   void Foo(this auto&& self)
   {
       std::cout << "Foo call\n";
   }
};

void test_TA()
{
    TA  a;
    void (*fun_ptr)(TA&) = &TA::Foo;
    (*fun_ptr)(a);
}

int main()
{
    test_TA();
}

This is somewhat counter-intuitive.
Why was this design adopted? Is there any difficulty for compilers in using the ordinary type of a pointer to a class method (void (TA::*mem_ptr)() &) in this case? It would be intuitive.