0.00/0.05	NO
0.00/0.05	(ignored inputs)COMMENT [75] Example 31 submitted by: Thomas Sternagel and Aart Middeldorp
0.00/0.05	Conditional Rewrite Rules:
0.00/0.05	   [ a -> c,
0.00/0.05	     a -> d,
0.00/0.05	     f(?x,?y) -> ?z | c == ?x,?y == ?z ]
0.00/0.05	Check whether all rules are type 3
0.00/0.05	OK
0.00/0.05	Check whether the input is deterministic
0.00/0.05	OK
0.00/0.05	Result of unraveling:
0.00/0.05	   [ a -> c,
0.00/0.05	     a -> d,
0.00/0.05	     f(?x,?y) -> U0(c,?x,?y),
0.00/0.05	     U0(?x,?x,?y) -> U1(?y,?y,?x),
0.00/0.05	     U1(?z,?y,?x) -> ?z ]
0.00/0.05	Check whether U(R) is terminating
0.00/0.05	OK
0.00/0.05	Check whether the input is weakly left-linear
0.00/0.05	OK
0.00/0.05	Check whether U(R) is confluent
0.00/0.05	U(R) is not confluent
0.00/0.05	Conditional critical pairs (CCPs):
0.00/0.05	   [ d = c,
0.00/0.05	     c = d ]
0.00/0.05	Check whether the input is almost orthogonale
0.00/0.05	not almost orthogonal
0.00/0.05	OK
0.00/0.05	Check whether the input is absolutely irreducible
0.00/0.05	OK
0.00/0.05	Check whether all CCPs are joinable
0.00/0.05	Some ccp may be feasible but not joinable
0.00/0.05	   [ d = c,
0.00/0.05	     c = d ]
0.00/0.05	A feasible but not joinable CCP: d = c
0.00/0.05	/export/starexec/sandbox/benchmark/theBenchmark.trs: Success(not CR)
0.00/0.05	(0 msec.)
0.00/0.05	EOF