mixin template question

[Paul D Anderson via Digitalmars-d-learn]

I don't understand why the following code compiles and runs 
without an error:

import std.stdio;
mixin template ABC(){
  int abc() { return 3; }
}

mixin ABC;

int abc() { return 4; }

void main()
{
  writefln("abc() = %s", abc());
}

Doesn't the mixin ABC create a function with the same signature 
as the "actual function" abc()?

It compiles with both included and writes "abc() = 4". If I 
comment out the actual function then it writes "abc() = 3". The 
actual function takes precedence, but why don't they conflict?

Paul