0.00/0.02	YES
0.00/0.02	(ignored inputs)COMMENT [75] Example 33 submitted by: Thomas Sternagel and Aart Middeldorp
0.00/0.02	Conditional Rewrite Rules:
0.00/0.02	   [ f(?x) -> ?x | ?x == a,
0.00/0.02	     g(?x) -> C | A == B,
0.00/0.02	     A -> B ]
0.00/0.02	Check whether all rules are type 3
0.00/0.02	OK
0.00/0.02	Check whether the input is deterministic
0.00/0.02	OK
0.00/0.02	Result of unraveling:
0.00/0.02	   [ f(?x) -> U0(?x,?x),
0.00/0.02	     U0(a,?x) -> ?x,
0.00/0.02	     g(?x) -> U1(A,?x),
0.00/0.02	     U1(B,?x) -> C,
0.00/0.02	     A -> B ]
0.00/0.02	Check whether U(R) is terminating
0.00/0.02	OK
0.00/0.02	Check whether the input is weakly left-linear
0.00/0.02	OK
0.00/0.02	Check whether U(R) is confluent
0.00/0.02	OK
0.00/0.02	R is deterministic, weakly left-linear and U(R) is confluent
0.00/0.02	/export/starexec/sandbox/benchmark/theBenchmark.trs: Success(CR)
0.00/0.02	(0 msec.)
0.00/0.02	EOF