0.00/0.02	YES
0.00/0.02	(ignored inputs)COMMENT [68] Example 3.2 doi:10.1007/3-540-59200-8_56 submitted by: Thomas Sternagel and Aart Middeldorp
0.00/0.02	Conditional Rewrite Rules:
0.00/0.02	   [ f(?x) -> g(?x,?y,?z) | h(a,?x) == i(?y),h(a,?y) == i(?z),
0.00/0.02	     h(a,a) -> i(b),
0.00/0.02	     h(a,b) -> i(c),
0.00/0.02	     h(b,b) -> i(d) ]
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(h(a,?x),?x),
0.00/0.02	     U0(i(?y),?x) -> U1(h(a,?y),?x,?y),
0.00/0.02	     U1(i(?z),?x,?y) -> g(?x,?y,?z),
0.00/0.02	     h(a,a) -> i(b),
0.00/0.02	     h(a,b) -> i(c),
0.00/0.02	     h(b,b) -> i(d) ]
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