Why is copy assigment possible, if a class has only a (templated) move assignment operator?

I have stumbled over code today, that I don't understand. Please consider the following example:

#include <iostream>
#include <string>

class A
{
public:
    template <class Type>
    Type& operator=(Type&& theOther)
    {
        text = std::forward<Type>(theOther).text;

        return *this;
    }

private:
    std::string text;
};

class B
{
public:
    B& operator=(B&& theOther)
    {
        text = std::forward<B>(theOther).text;

        return *this;
    }

private:
    std::string text;
};

int main()
{
    A a1;
    A a2;
    a2 = a1;

    B b1;
    B b2;
    b2 = b1;

    return 0;
}

When compiling, MinGW-w64/g++ 10.2 states:

..\src\Main.cpp: In function 'int main()':
..\src\Main.cpp:41:7: error: use of deleted function 'B& B::operator=(const B&)'
   41 |  b2 = b1;
      |       ^~
..\src\Main.cpp:19:7: note: 'B& B::operator=(const B&)' is implicitly declared as deleted because 'B' declares a move constructor or move assignment operator
   19 | class B
      |       ^
mingw32-make: *** [Makefile:419: Main.o] Error 1

I fully understand the error message. But I don't understand why I don't get the same message with class A. Isn't the templated move assignment operator also a move assignment operator? Why then is the copy assignment operator not deleted? Is this well-written code?