0.00/0.04	YES
0.00/0.04	(ignored inputs)COMMENT doi:10.1007/3-540-58216-9_40 [45] Example 2.1 ( b ) submitted by: Thomas Sternagel
0.00/0.04	Conditional Rewrite Rules:
0.00/0.04	   [ plus(0,?y) -> ?y,
0.00/0.04	     plus(s(?x),?y) -> s(plus(?x,?y)),
0.00/0.04	     fib(0) -> pair(s(0),0),
0.00/0.04	     fib(s(?x)) -> pair(?z3,?z1) | fib(?x) == pair(?z1,?z2),plus(?z1,?z2) == ?z3 ]
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	   [ plus(0,?y) -> ?y,
0.00/0.04	     plus(s(?x),?y) -> s(plus(?x,?y)),
0.00/0.04	     fib(0) -> pair(s(0),0),
0.00/0.04	     fib(s(?x)) -> U0(fib(?x),?x),
0.00/0.04	     U0(pair(?z1,?z2),?x) -> U1(plus(?z1,?z2),?x,?z1,?z2),
0.00/0.04	     U1(?z3,?x,?z1,?z2) -> pair(?z3,?z1) ]
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/sandbox/benchmark/theBenchmark.trs: Success(CR)
0.00/0.04	(8 msec.)
0.00/0.04	EOF