0.00/0.06 YES 0.00/0.06 (ignored inputs)COMMENT doi:10.1007/978-1-4757-3661-8 [18] p. 205 submitted by: Thomas Sternagel and Aart Middeldorp 0.00/0.06 Conditional Rewrite Rules: 0.00/0.06 [ le(0,?x) -> true, 0.00/0.06 le(s(?x),0) -> false, 0.00/0.06 le(s(?x),s(?y)) -> le(?x,?y), 0.00/0.06 app(nil,?x) -> ?x, 0.00/0.06 app(cons(?x,?xs),?ys) -> cons(?x,app(?xs,?ys)), 0.00/0.06 split(?x,nil) -> pair(nil,nil), 0.00/0.06 qsort(nil) -> nil, 0.00/0.06 split(?x,cons(?y,?ys)) -> pair(?xs,cons(?y,?zs)) | split(?x,?ys) == pair(?xs,?zs),le(?x,?y) == true, 0.00/0.06 split(?x,cons(?y,?ys)) -> pair(cons(?y,?xs),?zs) | split(?x,?ys) == pair(?xs,?zs),le(?x,?y) == false, 0.00/0.06 qsort(cons(?x,?xs)) -> app(qsort(?ys),cons(?x,qsort(?zs))) | split(?x,?xs) == pair(?ys,?zs) ] 0.00/0.06 Check whether all rules are type 3 0.00/0.06 OK 0.00/0.06 Check whether the input is deterministic 0.00/0.06 OK 0.00/0.06 Result of unraveling: 0.00/0.06 [ le(0,?x) -> true, 0.00/0.06 le(s(?x),0) -> false, 0.00/0.06 le(s(?x),s(?y)) -> le(?x,?y), 0.00/0.06 app(nil,?x) -> ?x, 0.00/0.06 app(cons(?x,?xs),?ys) -> cons(?x,app(?xs,?ys)), 0.00/0.06 split(?x,nil) -> pair(nil,nil), 0.00/0.06 qsort(nil) -> nil, 0.00/0.06 split(?x,cons(?y,?ys)) -> U2(split(?x,?ys),?x,?y,?ys), 0.00/0.06 U2(pair(?xs,?zs),?x,?y,?ys) -> U3(le(?x,?y),?x,?y,?ys,?xs,?zs), 0.00/0.06 U3(true,?x,?y,?ys,?xs,?zs) -> pair(?xs,cons(?y,?zs)), 0.00/0.06 U3(false,?x,?y,?ys,?xs,?zs) -> pair(cons(?y,?xs),?zs), 0.00/0.06 qsort(cons(?x,?xs)) -> U4(split(?x,?xs),?x,?xs), 0.00/0.06 U4(pair(?ys,?zs),?x,?xs) -> app(qsort(?ys),cons(?x,qsort(?zs))) ] 0.00/0.06 Check whether U(R) is terminating 0.00/0.06 OK 0.00/0.06 Check whether the input is weakly left-linear 0.00/0.06 OK 0.00/0.06 Check whether U(R) is confluent 0.00/0.06 OK 0.00/0.06 R is deterministic, weakly left-linear and U(R) is confluent 0.00/0.06 /export/starexec/sandbox/benchmark/theBenchmark.trs: Success(CR) 0.00/0.06 (20 msec.) 0.00/0.06 EOF