0.00/0.02	YES
0.00/0.02	(ignored inputs)COMMENT doi:10.2168/LMCS-8 ( 3:4 ) 2012 [64] Example 3.4 ( R_4 ) submitted by: Thomas Sternagel and Aart Middeldorp
0.00/0.02	Conditional Rewrite Rules:
0.00/0.02	   [ f(?x,?y) -> ?x | g(?x) == ?z,g(?y) == ?z,
0.00/0.02	     g(?x) -> c | d == c ]
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,?y) -> U0(g(?x),?x,?y),
0.00/0.02	     U0(?z,?x,?y) -> U1(g(?y),?x,?y,?z),
0.00/0.02	     U1(?z,?x,?y,?z) -> ?x,
0.00/0.02	     g(?x) -> U2(d,?x),
0.00/0.02	     U2(c,?x) -> c ]
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