NO
(ignored inputs)COMMENT [111] p. 27
Input:
   [ a -> b,
     f(?x,a) -> b,
     f(b,b) -> b ]

Make it flat:
   [ a -> b,
     f(?x,a) -> b,
     f(b,b) -> b ]
Time: 0.000 [s]

Make it Complete (R^):
   [ a = f(b,b),
     f(b,a) = b,
     f(a,b) = b,
     a = f(?x,a),
     f(b,a) = f(?x,a),
     f(a,b) = f(?x,a),
     f(?x,b) = b,
     f(?x_1,b) = f(?x,a),
     f(b,a) = f(?x,b),
     f(a,a) = f(?x,b),
     f(b,a) = f(a,b),
     f(b,a) = f(a,a),
     f(a,b) = f(a,a),
     f(b,a) = f(b,b),
     a = f(?x_1,b),
     f(a,b) = a,
     f(b,a) = a,
     a = f(a,a),
     f(a,a) = b,
     f(a,a) = f(?x_1,a),
     f(a,a) = f(b,b),
     f(a,b) = f(b,b),
     f(a,b) = f(?x,b),
     f(?x,b) = f(?x_1,b),
     f(?x,b) = f(b,b),
     a = b,
     f(?x,a) = f(b,b),
     f(?x,a) = f(?x_1,a),
     f(?x,a) = b,
     f(b,b) = b ]
Time: 0.034 [s]

The number of normal forms that must be checked: 845
Time: 0.252 [s]

Now checking all the pairs...

Time to check pairs: 0.000 [s]

The TRS doesn't have Uniqueness of Normal Forms.
Counter Example: 
     f(f(?!cx_2,?!cx_3),b)
<->* f(f(?!cx_3,?!cx_3),b)

proof:
     f(f(?!cx_2,?!cx_3),b)
->R^ f(f(?!cx_2,?!cx_3),a)
->R^ b
->R^ a

     f(f(?!cx_3,?!cx_3),b)
->R^ f(f(?!cx_3,?!cx_3),a)
->R^ b
->R^ a
Total Time: 0.287 [s]

problems/505.trs: Success(not UNC)
real 0.33
user 0.29
sys 0.02