To find out which equation is applicable, we have to calculate whether m0 contains the key k0:

      emptyMap().add("blue",1).add("green",2).containsKey("blue")
      
                m0            .add( k0, v0).containsKey( x )
    

There are two equations that might match that as well, so we have to calculate whether x.equals(k0), that is, whether "blue".equals("green"). Our knowledge of Java tells us that comes out false, so we simplify using the second equation for containsKey:

      emptyMap().add("blue",1).add("green",2).containsKey("blue")
      
                m0            .add( k0, v0).containsKey( x )
      
      =
      emptyMap().add("blue",1).containsKey("blue")
      
         m0     .add( k0, v0).containsKey("blue")