0.00/0.02 YES 0.00/0.02 (ignored inputs)COMMENT [121] Example 4.13 submitted by: Thomas Sternagel 0.00/0.02 Conditional Rewrite Rules: 0.00/0.02 [ f(?x) -> a | ?x == a, 0.00/0.02 f(?x) -> b | ?x == b ] 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) -> U1(?x,?x), 0.00/0.02 U1(a,?x) -> a, 0.00/0.02 U1(b,?x) -> b ] 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