0.00/0.04	YES
0.00/0.04	(ignored inputs)COMMENT doi:10.2168/LMCS-8 ( 3:4 ) 2012 [64] Example 4.4 ( R_7 ) submitted by: Thomas Sternagel and Aart Middeldorp
0.00/0.04	Conditional Rewrite Rules:
0.00/0.04	   [ split(?x,nil) -> tp2(nil,nil),
0.00/0.04	     split(?x,cons(?y,?ys)) -> tp2(?zs1,cons(?y,?zs2)) | split(?x,?ys) == tp2(?zs1,?zs2),le(?x,?y) == true,
0.00/0.04	     split(?x,cons(?y,?ys)) -> tp2(cons(?y,?zs1),?zs2) | split(?x,?ys) == tp2(?zs1,?zs2),le(?x,?y) == false,
0.00/0.04	     le(0,?y) -> true,
0.00/0.04	     le(s(?x),0) -> false,
0.00/0.04	     le(s(?x),s(?y)) -> le(?x,?y) ]
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	   [ split(?x,nil) -> tp2(nil,nil),
0.00/0.04	     split(?x,cons(?y,?ys)) -> U2(split(?x,?ys),?x,?y,?ys),
0.00/0.04	     U2(tp2(?zs1,?zs2),?x,?y,?ys) -> U3(le(?x,?y),?x,?y,?ys,?zs1,?zs2),
0.00/0.04	     U3(true,?x,?y,?ys,?zs1,?zs2) -> tp2(?zs1,cons(?y,?zs2)),
0.00/0.04	     U3(false,?x,?y,?ys,?zs1,?zs2) -> tp2(cons(?y,?zs1),?zs2),
0.00/0.04	     le(0,?y) -> true,
0.00/0.04	     le(s(?x),0) -> false,
0.00/0.04	     le(s(?x),s(?y)) -> le(?x,?y) ]
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	OK
0.00/0.04	R is deterministic, weakly left-linear and U(R) is confluent
0.00/0.04	/export/starexec/sandbox2/benchmark/theBenchmark.trs: Success(CR)
0.00/0.04	(8 msec.)
0.00/0.04	EOF