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


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


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 21 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) QDPOrderProof [EQUIVALENT, 177 ms]
        (27) QDP
        (28) QDPOrderProof [EQUIVALENT, 32 ms]
        (29) QDP
        (30) QDPOrderProof [EQUIVALENT, 179 ms]
        (31) QDP
        (32) QDPOrderProof [EQUIVALENT, 66 ms]
        (33) QDP
        (34) QDPOrderProof [EQUIVALENT, 42 ms]
        (35) QDP
        (36) QDPOrderProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) QDPOrderProof [EQUIVALENT, 0 ms]
        (39) QDP
        (40) QDPOrderProof [EQUIVALENT, 0 ms]
        (41) QDP
        (42) QDPOrderProof [EQUIVALENT, 43 ms]
        (43) QDP
        (44) QDPOrderProof [EQUIVALENT, 30 ms]
        (45) QDP
        (46) DependencyGraphProof [EQUIVALENT, 0 ms]
        (47) TRUE


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

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

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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(after(0, XS)) -> MARK(XS)
   ACTIVE(after(s(N), cons(X, XS))) -> MARK(after(N, XS))
   ACTIVE(after(s(N), cons(X, XS))) -> AFTER(N, XS)
   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(after(X1, X2)) -> ACTIVE(after(mark(X1), mark(X2)))
   MARK(after(X1, X2)) -> AFTER(mark(X1), mark(X2))
   MARK(after(X1, X2)) -> MARK(X1)
   MARK(after(X1, X2)) -> MARK(X2)
   MARK(0) -> ACTIVE(0)
   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)
   AFTER(mark(X1), X2) -> AFTER(X1, X2)
   AFTER(X1, mark(X2)) -> AFTER(X1, X2)
   AFTER(active(X1), X2) -> AFTER(X1, X2)
   AFTER(X1, active(X2)) -> AFTER(X1, X2)

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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 5 SCCs with 9 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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:

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

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


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


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


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


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

(9)
YES

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

(10)
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(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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:

   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.
----------------------------------------

(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:
*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


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

(14)
YES

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

(15)
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(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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:

   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.
----------------------------------------

(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:
*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


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

(19)
YES

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

(20)
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(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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:

   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.
----------------------------------------

(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:
*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


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

(24)
YES

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

(25)
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(after(0, XS)) -> MARK(XS)
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   ACTIVE(after(s(N), cons(X, XS))) -> MARK(after(N, XS))
   MARK(s(X)) -> MARK(X)
   MARK(after(X1, X2)) -> ACTIVE(after(mark(X1), mark(X2)))
   MARK(after(X1, X2)) -> MARK(X1)
   MARK(after(X1, X2)) -> MARK(X2)

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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

(26) 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

s(x1)  =  s

after(x1, x2)  =  after

0  =  0

active(x1)  =  x1


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

Status: MARK: multiset status
from: multiset status
cons: multiset status
s: multiset status
after: 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)
   after(X1, mark(X2)) -> after(X1, X2)
   after(mark(X1), X2) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)


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

(27)
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(after(0, XS)) -> MARK(XS)
   MARK(cons(X1, X2)) -> MARK(X1)
   ACTIVE(after(s(N), cons(X, XS))) -> MARK(after(N, XS))
   MARK(s(X)) -> MARK(X)
   MARK(after(X1, X2)) -> ACTIVE(after(mark(X1), mark(X2)))
   MARK(after(X1, X2)) -> MARK(X1)
   MARK(after(X1, X2)) -> MARK(X2)

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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

(28) 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:  Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(ACTIVE(x_1)) = x_1
   POL(MARK(x_1)) = 1
   POL(active(x_1)) = 0
   POL(after(x_1, x_2)) = 1
   POL(cons(x_1, x_2)) = 0
   POL(from(x_1)) = 1
   POL(mark(x_1)) = 0
   POL(s(x_1)) = 0

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


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

(29)
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(after(0, XS)) -> MARK(XS)
   MARK(cons(X1, X2)) -> MARK(X1)
   ACTIVE(after(s(N), cons(X, XS))) -> MARK(after(N, XS))
   MARK(s(X)) -> MARK(X)
   MARK(after(X1, X2)) -> ACTIVE(after(mark(X1), mark(X2)))
   MARK(after(X1, X2)) -> MARK(X1)
   MARK(after(X1, X2)) -> MARK(X2)

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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

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


The following pairs can be oriented strictly and are deleted.

   ACTIVE(after(0, XS)) -> MARK(XS)
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(MARK(x_1)) =  	[[0A]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(from(x_1)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(ACTIVE(x_1)) =  	[[0A]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(mark(x_1)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(cons(x_1, x_2)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_1 	 +  	[[0A]] 	* 	x_2
>>>

   <<<
 POL(s(x_1)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(after(x_1, x_2)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_1 	 +  	[[1A]] 	* 	x_2
>>>

   <<<
 POL(0) =  	[[2A]]
>>>

   <<<
 POL(active(x_1)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_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(after(0, XS)) -> mark(XS)
   mark(s(X)) -> active(s(mark(X)))
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   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)
   after(X1, mark(X2)) -> after(X1, X2)
   after(mark(X1), X2) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)


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

(31)
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)) -> MARK(X1)
   ACTIVE(after(s(N), cons(X, XS))) -> MARK(after(N, XS))
   MARK(s(X)) -> MARK(X)
   MARK(after(X1, X2)) -> ACTIVE(after(mark(X1), mark(X2)))
   MARK(after(X1, X2)) -> MARK(X1)
   MARK(after(X1, X2)) -> MARK(X2)

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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

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


The following pairs can be oriented strictly and are deleted.

   MARK(from(X)) -> MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(MARK(x_1)) =  	[[0A]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(from(x_1)) =  	[[1A]] 	 +  	[[1A]] 	* 	x_1
>>>

   <<<
 POL(ACTIVE(x_1)) =  	[[0A]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(mark(x_1)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(cons(x_1, x_2)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_1 	 +  	[[0A]] 	* 	x_2
>>>

   <<<
 POL(s(x_1)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(after(x_1, x_2)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_1 	 +  	[[0A]] 	* 	x_2
>>>

   <<<
 POL(active(x_1)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(0) =  	[[0A]]
>>>


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(after(0, XS)) -> mark(XS)
   mark(s(X)) -> active(s(mark(X)))
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   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)
   after(X1, mark(X2)) -> after(X1, X2)
   after(mark(X1), X2) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)


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

(33)
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(cons(X1, X2)) -> MARK(X1)
   ACTIVE(after(s(N), cons(X, XS))) -> MARK(after(N, XS))
   MARK(s(X)) -> MARK(X)
   MARK(after(X1, X2)) -> ACTIVE(after(mark(X1), mark(X2)))
   MARK(after(X1, X2)) -> MARK(X1)
   MARK(after(X1, X2)) -> MARK(X2)

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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

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


The following pairs can be oriented strictly and are deleted.

   MARK(cons(X1, X2)) -> MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(MARK(x_1)) =  	[[0A]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(from(x_1)) =  	[[1A]] 	 +  	[[1A]] 	* 	x_1
>>>

   <<<
 POL(ACTIVE(x_1)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(mark(x_1)) =  	[[0A]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(cons(x_1, x_2)) =  	[[1A]] 	 +  	[[1A]] 	* 	x_1 	 +  	[[0A]] 	* 	x_2
>>>

   <<<
 POL(s(x_1)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(after(x_1, x_2)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_1 	 +  	[[0A]] 	* 	x_2
>>>

   <<<
 POL(active(x_1)) =  	[[-I]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(0) =  	[[0A]]
>>>


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(after(0, XS)) -> mark(XS)
   mark(s(X)) -> active(s(mark(X)))
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   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)
   after(X1, mark(X2)) -> after(X1, X2)
   after(mark(X1), X2) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)


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

(35)
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))))
   ACTIVE(after(s(N), cons(X, XS))) -> MARK(after(N, XS))
   MARK(s(X)) -> MARK(X)
   MARK(after(X1, X2)) -> ACTIVE(after(mark(X1), mark(X2)))
   MARK(after(X1, X2)) -> MARK(X1)
   MARK(after(X1, X2)) -> MARK(X2)

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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

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


The following pairs can be oriented strictly and are deleted.

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

from(x1)  =  from

ACTIVE(x1)  =  x1

cons(x1, x2)  =  x2

after(x1, x2)  =  after(x1, x2)

s(x1)  =  x1

mark(x1)  =  x1

active(x1)  =  x1

0  =  0


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   0=1
   from=2
   after_2=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(after(0, XS)) -> mark(XS)
   mark(s(X)) -> active(s(mark(X)))
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   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)
   after(X1, mark(X2)) -> after(X1, X2)
   after(mark(X1), X2) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)


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

(37)
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))))
   ACTIVE(after(s(N), cons(X, XS))) -> MARK(after(N, XS))
   MARK(s(X)) -> MARK(X)
   MARK(after(X1, X2)) -> ACTIVE(after(mark(X1), mark(X2)))

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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

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


The following pairs can be oriented strictly and are deleted.

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

from(x1)  =  from

ACTIVE(x1)  =  x1

cons(x1, x2)  =  cons

after(x1, x2)  =  after

s(x1)  =  x1


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   from=2
   cons=1
   after=2

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


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

(39)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(from(X)) -> ACTIVE(from(mark(X)))
   ACTIVE(after(s(N), cons(X, XS))) -> MARK(after(N, XS))
   MARK(s(X)) -> MARK(X)
   MARK(after(X1, X2)) -> ACTIVE(after(mark(X1), mark(X2)))

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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

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


The following pairs can be oriented strictly and are deleted.

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

from(x1)  =  from

ACTIVE(x1)  =  x1

after(x1, x2)  =  after

s(x1)  =  s(x1)


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   s_1=1
   from=2
   after=1

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


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

(41)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(from(X)) -> ACTIVE(from(mark(X)))
   ACTIVE(after(s(N), cons(X, XS))) -> MARK(after(N, XS))
   MARK(after(X1, X2)) -> ACTIVE(after(mark(X1), mark(X2)))

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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

(42) 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)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( ACTIVE_1(x_1) ) = max{0, -2}
POL( after_2(x_1, x_2) ) = max{0, -2}
POL( from_1(x_1) ) = 2x_1 + 2
POL( mark_1(x_1) ) = 2
POL( active_1(x_1) ) = max{0, x_1 - 2}
POL( cons_2(x_1, x_2) ) = max{0, 2x_1 - 2}
POL( s_1(x_1) ) = x_1
POL( 0 ) = 0
POL( MARK_1(x_1) ) = max{0, 2x_1 - 2}

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

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


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

(43)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(after(s(N), cons(X, XS))) -> MARK(after(N, XS))
   MARK(after(X1, X2)) -> ACTIVE(after(mark(X1), mark(X2)))

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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

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


The following pairs can be oriented strictly and are deleted.

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

after(x1, x2)  =  after(x1, x2)

s(x1)  =  s(x1)

cons(x1, x2)  =  x2

MARK(x1)  =  MARK(x1)

mark(x1)  =  x1

active(x1)  =  x1

from(x1)  =  from

0  =  0


Recursive path order with status [RPO].
Quasi-Precedence: [ACTIVE_1, MARK_1] > after_2
from > s_1

Status: ACTIVE_1: multiset status
after_2: [1,2]
s_1: multiset status
MARK_1: multiset status
from: []
0: multiset status


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

   after(X1, mark(X2)) -> after(X1, X2)
   after(mark(X1), X2) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)
   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(after(0, XS)) -> mark(XS)
   mark(s(X)) -> active(s(mark(X)))
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   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)


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

(45)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(after(X1, X2)) -> ACTIVE(after(mark(X1), mark(X2)))

The TRS R consists of the following rules:

   active(from(X)) -> mark(cons(X, from(s(X))))
   active(after(0, XS)) -> mark(XS)
   active(after(s(N), cons(X, XS))) -> mark(after(N, XS))
   mark(from(X)) -> active(from(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(s(X)) -> active(s(mark(X)))
   mark(after(X1, X2)) -> active(after(mark(X1), mark(X2)))
   mark(0) -> active(0)
   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)
   after(mark(X1), X2) -> after(X1, X2)
   after(X1, mark(X2)) -> after(X1, X2)
   after(active(X1), X2) -> after(X1, X2)
   after(X1, active(X2)) -> after(X1, X2)

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

(46) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(47)
TRUE