/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 10 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) UsableRulesProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) UsableRulesProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES
    (25) QDP
        (26) UsableRulesProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (29) YES
    (30) QDP
        (31) QDPOrderProof [EQUIVALENT, 107 ms]
        (32) QDP
        (33) QDPOrderProof [EQUIVALENT, 135 ms]
        (34) QDP
        (35) QDPOrderProof [EQUIVALENT, 85 ms]
        (36) QDP
        (37) UsableRulesProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (40) QDP
        (41) DependencyGraphProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (44) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(length(nil)) -> mark(0)
   active(length(cons(X, Y))) -> mark(s(length1(Y)))
   active(length1(X)) -> mark(length(X))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(length(X)) -> active(length(X))
   mark(nil) -> active(nil)
   mark(0) -> active(0)
   mark(length1(X)) -> active(length1(X))
   from(mark(X)) -> from(X)
   from(active(X)) -> from(X)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   length(mark(X)) -> length(X)
   length(active(X)) -> length(X)
   length1(mark(X)) -> length1(X)
   length1(active(X)) -> length1(X)

Q is empty.

----------------------------------------

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(from(X)) -> MARK(cons(X, from(s(X))))
   ACTIVE(from(X)) -> CONS(X, from(s(X)))
   ACTIVE(from(X)) -> FROM(s(X))
   ACTIVE(from(X)) -> S(X)
   ACTIVE(length(nil)) -> MARK(0)
   ACTIVE(length(cons(X, Y))) -> MARK(s(length1(Y)))
   ACTIVE(length(cons(X, Y))) -> S(length1(Y))
   ACTIVE(length(cons(X, Y))) -> LENGTH1(Y)
   ACTIVE(length1(X)) -> MARK(length(X))
   ACTIVE(length1(X)) -> LENGTH(X)
   MARK(from(X)) -> ACTIVE(from(mark(X)))
   MARK(from(X)) -> FROM(mark(X))
   MARK(from(X)) -> MARK(X)
   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   MARK(cons(X1, X2)) -> CONS(mark(X1), X2)
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(s(X)) -> S(mark(X))
   MARK(s(X)) -> MARK(X)
   MARK(length(X)) -> ACTIVE(length(X))
   MARK(nil) -> ACTIVE(nil)
   MARK(0) -> ACTIVE(0)
   MARK(length1(X)) -> ACTIVE(length1(X))
   FROM(mark(X)) -> FROM(X)
   FROM(active(X)) -> FROM(X)
   CONS(mark(X1), X2) -> CONS(X1, X2)
   CONS(X1, mark(X2)) -> CONS(X1, X2)
   CONS(active(X1), X2) -> CONS(X1, X2)
   CONS(X1, active(X2)) -> CONS(X1, X2)
   S(mark(X)) -> S(X)
   S(active(X)) -> S(X)
   LENGTH(mark(X)) -> LENGTH(X)
   LENGTH(active(X)) -> LENGTH(X)
   LENGTH1(mark(X)) -> LENGTH1(X)
   LENGTH1(active(X)) -> LENGTH1(X)

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(length(nil)) -> mark(0)
   active(length(cons(X, Y))) -> mark(s(length1(Y)))
   active(length1(X)) -> mark(length(X))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(length(X)) -> active(length(X))
   mark(nil) -> active(nil)
   mark(0) -> active(0)
   mark(length1(X)) -> active(length1(X))
   from(mark(X)) -> from(X)
   from(active(X)) -> from(X)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   length(mark(X)) -> length(X)
   length(active(X)) -> length(X)
   length1(mark(X)) -> length1(X)
   length1(active(X)) -> length1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 12 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

----------------------------------------

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LENGTH1(active(X)) -> LENGTH1(X)
   LENGTH1(mark(X)) -> LENGTH1(X)

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(length(nil)) -> mark(0)
   active(length(cons(X, Y))) -> mark(s(length1(Y)))
   active(length1(X)) -> mark(length(X))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(length(X)) -> active(length(X))
   mark(nil) -> active(nil)
   mark(0) -> active(0)
   mark(length1(X)) -> active(length1(X))
   from(mark(X)) -> from(X)
   from(active(X)) -> from(X)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   length(mark(X)) -> length(X)
   length(active(X)) -> length(X)
   length1(mark(X)) -> length1(X)
   length1(active(X)) -> length1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LENGTH1(active(X)) -> LENGTH1(X)
   LENGTH1(mark(X)) -> LENGTH1(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(8) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*LENGTH1(active(X)) -> LENGTH1(X)
The graph contains the following edges 1 > 1


*LENGTH1(mark(X)) -> LENGTH1(X)
The graph contains the following edges 1 > 1


----------------------------------------

(9)
YES

----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LENGTH(active(X)) -> LENGTH(X)
   LENGTH(mark(X)) -> LENGTH(X)

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(length(nil)) -> mark(0)
   active(length(cons(X, Y))) -> mark(s(length1(Y)))
   active(length1(X)) -> mark(length(X))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(length(X)) -> active(length(X))
   mark(nil) -> active(nil)
   mark(0) -> active(0)
   mark(length1(X)) -> active(length1(X))
   from(mark(X)) -> from(X)
   from(active(X)) -> from(X)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   length(mark(X)) -> length(X)
   length(active(X)) -> length(X)
   length1(mark(X)) -> length1(X)
   length1(active(X)) -> length1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(11) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   LENGTH(active(X)) -> LENGTH(X)
   LENGTH(mark(X)) -> LENGTH(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(13) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*LENGTH(active(X)) -> LENGTH(X)
The graph contains the following edges 1 > 1


*LENGTH(mark(X)) -> LENGTH(X)
The graph contains the following edges 1 > 1


----------------------------------------

(14)
YES

----------------------------------------

(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(active(X)) -> S(X)
   S(mark(X)) -> S(X)

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(length(nil)) -> mark(0)
   active(length(cons(X, Y))) -> mark(s(length1(Y)))
   active(length1(X)) -> mark(length(X))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(length(X)) -> active(length(X))
   mark(nil) -> active(nil)
   mark(0) -> active(0)
   mark(length1(X)) -> active(length1(X))
   from(mark(X)) -> from(X)
   from(active(X)) -> from(X)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   length(mark(X)) -> length(X)
   length(active(X)) -> length(X)
   length1(mark(X)) -> length1(X)
   length1(active(X)) -> length1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(16) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(17)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   S(active(X)) -> S(X)
   S(mark(X)) -> S(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(18) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*S(active(X)) -> S(X)
The graph contains the following edges 1 > 1


*S(mark(X)) -> S(X)
The graph contains the following edges 1 > 1


----------------------------------------

(19)
YES

----------------------------------------

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   CONS(X1, mark(X2)) -> CONS(X1, X2)
   CONS(mark(X1), X2) -> CONS(X1, X2)
   CONS(active(X1), X2) -> CONS(X1, X2)
   CONS(X1, active(X2)) -> CONS(X1, X2)

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(length(nil)) -> mark(0)
   active(length(cons(X, Y))) -> mark(s(length1(Y)))
   active(length1(X)) -> mark(length(X))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(length(X)) -> active(length(X))
   mark(nil) -> active(nil)
   mark(0) -> active(0)
   mark(length1(X)) -> active(length1(X))
   from(mark(X)) -> from(X)
   from(active(X)) -> from(X)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   length(mark(X)) -> length(X)
   length(active(X)) -> length(X)
   length1(mark(X)) -> length1(X)
   length1(active(X)) -> length1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(21) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   CONS(X1, mark(X2)) -> CONS(X1, X2)
   CONS(mark(X1), X2) -> CONS(X1, X2)
   CONS(active(X1), X2) -> CONS(X1, X2)
   CONS(X1, active(X2)) -> CONS(X1, X2)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(23) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*CONS(X1, mark(X2)) -> CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2


*CONS(mark(X1), X2) -> CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2


*CONS(active(X1), X2) -> CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2


*CONS(X1, active(X2)) -> CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2


----------------------------------------

(24)
YES

----------------------------------------

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FROM(active(X)) -> FROM(X)
   FROM(mark(X)) -> FROM(X)

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(length(nil)) -> mark(0)
   active(length(cons(X, Y))) -> mark(s(length1(Y)))
   active(length1(X)) -> mark(length(X))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(length(X)) -> active(length(X))
   mark(nil) -> active(nil)
   mark(0) -> active(0)
   mark(length1(X)) -> active(length1(X))
   from(mark(X)) -> from(X)
   from(active(X)) -> from(X)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   length(mark(X)) -> length(X)
   length(active(X)) -> length(X)
   length1(mark(X)) -> length1(X)
   length1(active(X)) -> length1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(26) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(27)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FROM(active(X)) -> FROM(X)
   FROM(mark(X)) -> FROM(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(28) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*FROM(active(X)) -> FROM(X)
The graph contains the following edges 1 > 1


*FROM(mark(X)) -> FROM(X)
The graph contains the following edges 1 > 1


----------------------------------------

(29)
YES

----------------------------------------

(30)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(from(X)) -> ACTIVE(from(mark(X)))
   ACTIVE(from(X)) -> MARK(cons(X, from(s(X))))
   MARK(from(X)) -> MARK(X)
   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   ACTIVE(length(cons(X, Y))) -> MARK(s(length1(Y)))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   ACTIVE(length1(X)) -> MARK(length(X))
   MARK(s(X)) -> MARK(X)
   MARK(length(X)) -> ACTIVE(length(X))
   MARK(length1(X)) -> ACTIVE(length1(X))

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(length(nil)) -> mark(0)
   active(length(cons(X, Y))) -> mark(s(length1(Y)))
   active(length1(X)) -> mark(length(X))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(length(X)) -> active(length(X))
   mark(nil) -> active(nil)
   mark(0) -> active(0)
   mark(length1(X)) -> active(length1(X))
   from(mark(X)) -> from(X)
   from(active(X)) -> from(X)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   length(mark(X)) -> length(X)
   length(active(X)) -> length(X)
   length1(mark(X)) -> length1(X)
   length1(active(X)) -> length1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(31) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
MARK(x1)  =  MARK

from(x1)  =  from

ACTIVE(x1)  =  x1

mark(x1)  =  x1

cons(x1, x2)  =  cons

s(x1)  =  s

length(x1)  =  length

length1(x1)  =  length1

active(x1)  =  x1

nil  =  nil

0  =  0


Recursive path order with status [RPO].
Quasi-Precedence: [MARK, from, s, length, length1, 0] > cons

Status: MARK: multiset status
from: multiset status
cons: multiset status
s: multiset status
length: multiset status
length1: multiset status
nil: multiset status
0: multiset status


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   from(active(X)) -> from(X)
   from(mark(X)) -> from(X)
   s(active(X)) -> s(X)
   s(mark(X)) -> s(X)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   length1(active(X)) -> length1(X)
   length1(mark(X)) -> length1(X)
   length(active(X)) -> length(X)
   length(mark(X)) -> length(X)


----------------------------------------

(32)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(from(X)) -> ACTIVE(from(mark(X)))
   ACTIVE(from(X)) -> MARK(cons(X, from(s(X))))
   MARK(from(X)) -> MARK(X)
   ACTIVE(length(cons(X, Y))) -> MARK(s(length1(Y)))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   ACTIVE(length1(X)) -> MARK(length(X))
   MARK(s(X)) -> MARK(X)
   MARK(length(X)) -> ACTIVE(length(X))
   MARK(length1(X)) -> ACTIVE(length1(X))

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(length(nil)) -> mark(0)
   active(length(cons(X, Y))) -> mark(s(length1(Y)))
   active(length1(X)) -> mark(length(X))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(length(X)) -> active(length(X))
   mark(nil) -> active(nil)
   mark(0) -> active(0)
   mark(length1(X)) -> active(length1(X))
   from(mark(X)) -> from(X)
   from(active(X)) -> from(X)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   length(mark(X)) -> length(X)
   length(active(X)) -> length(X)
   length1(mark(X)) -> length1(X)
   length1(active(X)) -> length1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(33) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   MARK(s(X)) -> ACTIVE(s(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
MARK(x1)  =  MARK

from(x1)  =  from

ACTIVE(x1)  =  x1

mark(x1)  =  x1

cons(x1, x2)  =  cons(x1)

s(x1)  =  s

length(x1)  =  length

length1(x1)  =  length1

active(x1)  =  x1

nil  =  nil

0  =  0


Recursive path order with status [RPO].
Quasi-Precedence: [MARK, from, length, length1] > cons_1 > s
[MARK, from, length, length1] > 0

Status: MARK: multiset status
from: multiset status
cons_1: multiset status
s: multiset status
length: multiset status
length1: multiset status
nil: multiset status
0: multiset status


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   from(active(X)) -> from(X)
   from(mark(X)) -> from(X)
   s(active(X)) -> s(X)
   s(mark(X)) -> s(X)
   length1(active(X)) -> length1(X)
   length1(mark(X)) -> length1(X)
   length(active(X)) -> length(X)
   length(mark(X)) -> length(X)


----------------------------------------

(34)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(from(X)) -> ACTIVE(from(mark(X)))
   ACTIVE(from(X)) -> MARK(cons(X, from(s(X))))
   MARK(from(X)) -> MARK(X)
   ACTIVE(length(cons(X, Y))) -> MARK(s(length1(Y)))
   MARK(cons(X1, X2)) -> MARK(X1)
   ACTIVE(length1(X)) -> MARK(length(X))
   MARK(s(X)) -> MARK(X)
   MARK(length(X)) -> ACTIVE(length(X))
   MARK(length1(X)) -> ACTIVE(length1(X))

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(length(nil)) -> mark(0)
   active(length(cons(X, Y))) -> mark(s(length1(Y)))
   active(length1(X)) -> mark(length(X))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(length(X)) -> active(length(X))
   mark(nil) -> active(nil)
   mark(0) -> active(0)
   mark(length1(X)) -> active(length1(X))
   from(mark(X)) -> from(X)
   from(active(X)) -> from(X)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   length(mark(X)) -> length(X)
   length(active(X)) -> length(X)
   length1(mark(X)) -> length1(X)
   length1(active(X)) -> length1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(35) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   MARK(from(X)) -> ACTIVE(from(mark(X)))
   ACTIVE(from(X)) -> MARK(cons(X, from(s(X))))
   MARK(from(X)) -> MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( ACTIVE_1(x_1) ) = max{0, 2x_1 - 2}
POL( from_1(x_1) ) = x_1 + 2
POL( mark_1(x_1) ) = x_1
POL( active_1(x_1) ) = x_1
POL( cons_2(x_1, x_2) ) = x_1
POL( s_1(x_1) ) = 2x_1
POL( length_1(x_1) ) = 0
POL( length1_1(x_1) ) = 0
POL( nil ) = 0
POL( 0 ) = 0
POL( MARK_1(x_1) ) = 2x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   mark(from(X)) -> active(from(mark(X)))
   active(from(X)) -> mark(cons(X, from(s(X))))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   active(length(cons(X, Y))) -> mark(s(length1(Y)))
   mark(s(X)) -> active(s(mark(X)))
   active(length1(X)) -> mark(length(X))
   mark(length(X)) -> active(length(X))
   mark(length1(X)) -> active(length1(X))
   mark(nil) -> active(nil)
   mark(0) -> active(0)
   from(active(X)) -> from(X)
   from(mark(X)) -> from(X)
   s(active(X)) -> s(X)
   s(mark(X)) -> s(X)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   length1(active(X)) -> length1(X)
   length1(mark(X)) -> length1(X)
   length(active(X)) -> length(X)
   length(mark(X)) -> length(X)
   active(length(nil)) -> mark(0)


----------------------------------------

(36)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(length(cons(X, Y))) -> MARK(s(length1(Y)))
   MARK(cons(X1, X2)) -> MARK(X1)
   ACTIVE(length1(X)) -> MARK(length(X))
   MARK(s(X)) -> MARK(X)
   MARK(length(X)) -> ACTIVE(length(X))
   MARK(length1(X)) -> ACTIVE(length1(X))

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(length(nil)) -> mark(0)
   active(length(cons(X, Y))) -> mark(s(length1(Y)))
   active(length1(X)) -> mark(length(X))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(length(X)) -> active(length(X))
   mark(nil) -> active(nil)
   mark(0) -> active(0)
   mark(length1(X)) -> active(length1(X))
   from(mark(X)) -> from(X)
   from(active(X)) -> from(X)
   cons(mark(X1), X2) -> cons(X1, X2)
   cons(X1, mark(X2)) -> cons(X1, X2)
   cons(active(X1), X2) -> cons(X1, X2)
   cons(X1, active(X2)) -> cons(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(X)
   length(mark(X)) -> length(X)
   length(active(X)) -> length(X)
   length1(mark(X)) -> length1(X)
   length1(active(X)) -> length1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(37) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(38)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(length(cons(X, Y))) -> MARK(s(length1(Y)))
   MARK(cons(X1, X2)) -> MARK(X1)
   ACTIVE(length1(X)) -> MARK(length(X))
   MARK(s(X)) -> MARK(X)
   MARK(length(X)) -> ACTIVE(length(X))
   MARK(length1(X)) -> ACTIVE(length1(X))

The TRS R consists of the following rules:

   length1(active(X)) -> length1(X)
   length1(mark(X)) -> length1(X)
   length(active(X)) -> length(X)
   length(mark(X)) -> length(X)
   s(active(X)) -> s(X)
   s(mark(X)) -> s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(39) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

   ACTIVE(length(cons(X, Y))) -> MARK(s(length1(Y)))
   MARK(cons(X1, X2)) -> MARK(X1)
The following rules are removed from R:

   length1(active(X)) -> length1(X)
   length1(mark(X)) -> length1(X)
   length(active(X)) -> length(X)
   length(mark(X)) -> length(X)
   s(active(X)) -> s(X)
   s(mark(X)) -> s(X)
Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(ACTIVE(x_1)) = x_1
   POL(MARK(x_1)) = x_1
   POL(active(x_1)) = 2*x_1
   POL(cons(x_1, x_2)) = 1 + 2*x_1 + 2*x_2
   POL(length(x_1)) = 2*x_1
   POL(length1(x_1)) = 2*x_1
   POL(mark(x_1)) = 2*x_1
   POL(s(x_1)) = x_1


----------------------------------------

(40)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(length1(X)) -> MARK(length(X))
   MARK(s(X)) -> MARK(X)
   MARK(length(X)) -> ACTIVE(length(X))
   MARK(length1(X)) -> ACTIVE(length1(X))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(41) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

(42)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(s(X)) -> MARK(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(43) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*MARK(s(X)) -> MARK(X)
The graph contains the following edges 1 > 1


----------------------------------------

(44)
YES