Term Rewriting System R:
[X, Y, Z, X1, X2]
active(2nd(cons1(X, cons(Y, Z)))) -> mark(Y)
active(2nd(cons(X, X1))) -> mark(2nd(cons1(X, X1)))
active(from(X)) -> mark(cons(X, from(s(X))))
active(2nd(X)) -> 2nd(active(X))
active(cons(X1, X2)) -> cons(active(X1), X2)
active(from(X)) -> from(active(X))
active(s(X)) -> s(active(X))
active(cons1(X1, X2)) -> cons1(active(X1), X2)
active(cons1(X1, X2)) -> cons1(X1, active(X2))
2nd(mark(X)) -> mark(2nd(X))
2nd(ok(X)) -> ok(2nd(X))
cons(mark(X1), X2) -> mark(cons(X1, X2))
cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
from(mark(X)) -> mark(from(X))
from(ok(X)) -> ok(from(X))
s(mark(X)) -> mark(s(X))
s(ok(X)) -> ok(s(X))
cons1(mark(X1), X2) -> mark(cons1(X1, X2))
cons1(X1, mark(X2)) -> mark(cons1(X1, X2))
cons1(ok(X1), ok(X2)) -> ok(cons1(X1, X2))
proper(2nd(X)) -> 2nd(proper(X))
proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
proper(from(X)) -> from(proper(X))
proper(s(X)) -> s(proper(X))
proper(cons1(X1, X2)) -> cons1(proper(X1), proper(X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

ACTIVE(2nd(cons(X, X1))) -> 2ND(cons1(X, X1))
ACTIVE(2nd(cons(X, X1))) -> CONS1(X, X1)
ACTIVE(from(X)) -> CONS(X, from(s(X)))
ACTIVE(from(X)) -> FROM(s(X))
ACTIVE(from(X)) -> S(X)
ACTIVE(2nd(X)) -> 2ND(active(X))
ACTIVE(2nd(X)) -> ACTIVE(X)
ACTIVE(cons(X1, X2)) -> CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) -> ACTIVE(X1)
ACTIVE(from(X)) -> FROM(active(X))
ACTIVE(from(X)) -> ACTIVE(X)
ACTIVE(s(X)) -> S(active(X))
ACTIVE(s(X)) -> ACTIVE(X)
ACTIVE(cons1(X1, X2)) -> CONS1(active(X1), X2)
ACTIVE(cons1(X1, X2)) -> ACTIVE(X1)
ACTIVE(cons1(X1, X2)) -> CONS1(X1, active(X2))
ACTIVE(cons1(X1, X2)) -> ACTIVE(X2)
2ND(mark(X)) -> 2ND(X)
2ND(ok(X)) -> 2ND(X)
CONS(mark(X1), X2) -> CONS(X1, X2)
CONS(ok(X1), ok(X2)) -> CONS(X1, X2)
FROM(mark(X)) -> FROM(X)
FROM(ok(X)) -> FROM(X)
S(mark(X)) -> S(X)
S(ok(X)) -> S(X)
CONS1(mark(X1), X2) -> CONS1(X1, X2)
CONS1(X1, mark(X2)) -> CONS1(X1, X2)
CONS1(ok(X1), ok(X2)) -> CONS1(X1, X2)
PROPER(2nd(X)) -> 2ND(proper(X))
PROPER(2nd(X)) -> PROPER(X)
PROPER(cons(X1, X2)) -> CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) -> PROPER(X1)
PROPER(cons(X1, X2)) -> PROPER(X2)
PROPER(from(X)) -> FROM(proper(X))
PROPER(from(X)) -> PROPER(X)
PROPER(s(X)) -> S(proper(X))
PROPER(s(X)) -> PROPER(X)
PROPER(cons1(X1, X2)) -> CONS1(proper(X1), proper(X2))
PROPER(cons1(X1, X2)) -> PROPER(X1)
PROPER(cons1(X1, X2)) -> PROPER(X2)
TOP(mark(X)) -> TOP(proper(X))
TOP(mark(X)) -> PROPER(X)
TOP(ok(X)) -> TOP(active(X))
TOP(ok(X)) -> ACTIVE(X)

Furthermore, R contains eight SCCs.


   R
DPs
       →DP Problem 1
Remaining Obligation(s)
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)
       →DP Problem 6
Remaining Obligation(s)
       →DP Problem 7
Remaining Obligation(s)
       →DP Problem 8
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Remaining Obligation(s)
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)
       →DP Problem 6
Remaining Obligation(s)
       →DP Problem 7
Remaining Obligation(s)
       →DP Problem 8
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Remaining Obligation(s)
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)
       →DP Problem 6
Remaining Obligation(s)
       →DP Problem 7
Remaining Obligation(s)
       →DP Problem 8
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Remaining Obligation(s)
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)
       →DP Problem 6
Remaining Obligation(s)
       →DP Problem 7
Remaining Obligation(s)
       →DP Problem 8
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Remaining Obligation(s)
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)
       →DP Problem 6
Remaining Obligation(s)
       →DP Problem 7
Remaining Obligation(s)
       →DP Problem 8
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Remaining Obligation(s)
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)
       →DP Problem 6
Remaining Obligation(s)
       →DP Problem 7
Remaining Obligation(s)
       →DP Problem 8
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Remaining Obligation(s)
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)
       →DP Problem 6
Remaining Obligation(s)
       →DP Problem 7
Remaining Obligation(s)
       →DP Problem 8
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Remaining Obligation(s)
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)
       →DP Problem 6
Remaining Obligation(s)
       →DP Problem 7
Remaining Obligation(s)
       →DP Problem 8
Remaining Obligation(s)




The following remains to be proven:

Termination of R could not be shown.
Duration:
0:00 minutes