0.00/0.04	NO
0.00/0.04	(ignored inputs)COMMENT [75] Example 6 submitted by: Thomas Sternagel and Aart Middeldorp
0.00/0.04	Conditional Rewrite Rules:
0.00/0.04	   [ a -> c,
0.00/0.04	     a -> d,
0.00/0.04	     s(c) -> t(k),
0.00/0.04	     f(?x) -> ?z | s(?x) == t(?z) ]
0.00/0.04	Check whether all rules are type 3
0.00/0.04	OK
0.00/0.04	Check whether the input is deterministic
0.00/0.04	OK
0.00/0.04	Result of unraveling:
0.00/0.04	   [ a -> c,
0.00/0.04	     a -> d,
0.00/0.04	     s(c) -> t(k),
0.00/0.04	     f(?x) -> U0(s(?x),?x),
0.00/0.04	     U0(t(?z),?x) -> ?z ]
0.00/0.04	Check whether U(R) is terminating
0.00/0.04	OK
0.00/0.04	Check whether the input is weakly left-linear
0.00/0.04	OK
0.00/0.04	Check whether U(R) is confluent
0.00/0.04	U(R) is not confluent
0.00/0.04	Conditional critical pairs (CCPs):
0.00/0.04	   [ d = c,
0.00/0.04	     c = d ]
0.00/0.04	Check whether the input is almost orthogonale
0.00/0.04	not almost orthogonal
0.00/0.04	OK
0.00/0.04	Check whether the input is absolutely irreducible
0.00/0.04	OK
0.00/0.04	Check whether all CCPs are joinable
0.00/0.04	Some ccp may be feasible but not joinable
0.00/0.04	   [ d = c,
0.00/0.04	     c = d ]
0.00/0.04	A feasible but not joinable CCP: d = c
0.00/0.04	/export/starexec/sandbox/benchmark/theBenchmark.trs: Success(not CR)
0.00/0.04	(0 msec.)
0.00/0.05	EOF