Namespace qualified overloading of 'operator=='

I'm curious about why the following doesn't compile:

#include <iostream>
#include <functional>

namespace Bar {
struct Foo {
  int x;
};
}  // Namespace                                                                                                                     

static bool operator==(const Bar::Foo& a, const Bar::Foo& b) {
  return a.x == b.x;
}

int main() {
  Bar::Foo a = { 0 };
  Bar::Foo b = { 1 };

  // The following line is OK
  std::cout << (a == b) << std::endl;

  // The following line is not OK
  std::cout << std::equal_to<Bar::Foo>()(a, b) << std::endl;
}

The compiler barfs in this case:

[test]$ g++ --std=c++11 -o test test.cc
In file included from /usr/include/c++/4.8/string:48:0,
                 from /usr/include/c++/4.8/bits/locale_classes.h:40,
                 from /usr/include/c++/4.8/bits/ios_base.h:41,
                 from /usr/include/c++/4.8/ios:42,
                 from /usr/include/c++/4.8/ostream:38,
                 from /usr/include/c++/4.8/iostream:39,
                 from test.cc:1:
/usr/include/c++/4.8/bits/stl_function.h: In instantiation of ‘bool std::equal_to<_Tp>::operator()(const _Tp&, const _Tp&) const [with _Tp = Bar::Foo]’:
test.cc:18:46:   required from here
/usr/include/c++/4.8/bits/stl_function.h:208:20: error: no match for ‘operator==’ (operand types are ‘const Bar::Foo’ and ‘const Bar::Foo’)
       { return __x == __y; }
                    ^
/usr/include/c++/4.8/bits/stl_function.h:208:20: note: candidates are:
...

The offending line doesn't compile even if I try another variant:

namespace Bar {
struct Foo {
  int x;
  bool operator==(const Foo& o) {
    return x == o.x;
  }
};
}  // Namespace

However, it seems like the following does compile:

namespace Bar {
struct Foo {
  int x;
};

static bool operator==(const Foo& a, const Foo& b) {
  return a.x == b.x;
}
}  // Namespace

What the heck is going on here?