YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 3 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QReductionProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) QReductionProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES
    (19) QDP
        (20) UsableRulesProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) QReductionProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (25) YES
    (26) QDP
        (27) UsableRulesProof [EQUIVALENT, 0 ms]
        (28) QDP
        (29) QReductionProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (32) YES
    (33) QDP
        (34) QDPOrderProof [EQUIVALENT, 62 ms]
        (35) QDP
        (36) QDPOrderProof [EQUIVALENT, 17 ms]
        (37) QDP
        (38) QDPOrderProof [EQUIVALENT, 0 ms]
        (39) QDP
        (40) DependencyGraphProof [EQUIVALENT, 0 ms]
        (41) QDP
        (42) UsableRulesProof [EQUIVALENT, 0 ms]
        (43) QDP
        (44) QReductionProof [EQUIVALENT, 0 ms]
        (45) QDP
        (46) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (47) YES


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

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

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

The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))


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

(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(first(0, X)) -> MARK(nil)
   ACTIVE(first(s(X), cons(Y, Z))) -> MARK(cons(Y, first(X, Z)))
   ACTIVE(first(s(X), cons(Y, Z))) -> CONS(Y, first(X, Z))
   ACTIVE(first(s(X), cons(Y, Z))) -> FIRST(X, Z)
   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)
   MARK(first(X1, X2)) -> ACTIVE(first(mark(X1), mark(X2)))
   MARK(first(X1, X2)) -> FIRST(mark(X1), mark(X2))
   MARK(first(X1, X2)) -> MARK(X1)
   MARK(first(X1, X2)) -> MARK(X2)
   MARK(0) -> ACTIVE(0)
   MARK(nil) -> ACTIVE(nil)
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(s(X)) -> S(mark(X))
   MARK(s(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(from(X)) -> ACTIVE(from(mark(X)))
   MARK(from(X)) -> FROM(mark(X))
   MARK(from(X)) -> MARK(X)
   FIRST(mark(X1), X2) -> FIRST(X1, X2)
   FIRST(X1, mark(X2)) -> FIRST(X1, X2)
   FIRST(active(X1), X2) -> FIRST(X1, X2)
   FIRST(X1, active(X2)) -> FIRST(X1, X2)
   S(mark(X)) -> S(X)
   S(active(X)) -> S(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)
   FROM(mark(X)) -> FROM(X)
   FROM(active(X)) -> FROM(X)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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 12 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

(5)
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(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   active(from(X)) -> mark(cons(X, from(s(X))))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(0) -> active(0)
   mark(nil) -> active(nil)
   mark(s(X)) -> active(s(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(from(X)) -> active(from(mark(X)))
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(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)
   from(mark(X)) -> from(X)
   from(active(X)) -> from(X)

The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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

(6) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(7)
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.
The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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

(8) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))


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

(9)
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.
The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))

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

(10) 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


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

(11)
YES

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

(12)
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(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   active(from(X)) -> mark(cons(X, from(s(X))))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(0) -> active(0)
   mark(nil) -> active(nil)
   mark(s(X)) -> active(s(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(from(X)) -> active(from(mark(X)))
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(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)
   from(mark(X)) -> from(X)
   from(active(X)) -> from(X)

The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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

(13) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(14)
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.
The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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

(15) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))


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

(16)
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.
The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))

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

(17) 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


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

(18)
YES

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

(19)
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(first(0, X)) -> mark(nil)
   active(first(s(X), cons(Y, Z))) -> mark(cons(Y, first(X, Z)))
   active(from(X)) -> mark(cons(X, from(s(X))))
   mark(first(X1, X2)) -> active(first(mark(X1), mark(X2)))
   mark(0) -> active(0)
   mark(nil) -> active(nil)
   mark(s(X)) -> active(s(mark(X)))
   mark(cons(X1, X2)) -> active(cons(mark(X1), X2))
   mark(from(X)) -> active(from(mark(X)))
   first(mark(X1), X2) -> first(X1, X2)
   first(X1, mark(X2)) -> first(X1, X2)
   first(active(X1), X2) -> first(X1, X2)
   first(X1, active(X2)) -> first(X1, X2)
   s(mark(X)) -> s(X)
   s(active(X)) -> s(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)
   from(mark(X)) -> from(X)
   from(active(X)) -> from(X)

The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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

(20) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(21)
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.
The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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

(22) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))


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

(23)
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.
The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))

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

(24) 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


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

(25)
YES

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

(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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

(27) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

R is empty.
The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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

(29) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))


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

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

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

R is empty.
The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))

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

(31) 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:
*FIRST(X1, mark(X2)) -> FIRST(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2


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


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


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


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

(32)
YES

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

(33)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(first(s(X), cons(Y, Z))) -> MARK(cons(Y, first(X, Z)))
   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   ACTIVE(from(X)) -> MARK(cons(X, from(s(X))))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(first(X1, X2)) -> ACTIVE(first(mark(X1), mark(X2)))
   MARK(first(X1, X2)) -> MARK(X1)
   MARK(first(X1, X2)) -> MARK(X2)
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(s(X)) -> MARK(X)
   MARK(from(X)) -> ACTIVE(from(mark(X)))
   MARK(from(X)) -> MARK(X)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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.

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

first(x1, x2)  =  first(x1, x2)

s(x1)  =  x1

cons(x1, x2)  =  x1

MARK(x1)  =  x1

mark(x1)  =  x1

from(x1)  =  x1

active(x1)  =  x1

0  =  0

nil  =  nil


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   0=2
   first_2=2
   nil=4

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

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


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

(35)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   ACTIVE(from(X)) -> MARK(cons(X, from(s(X))))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(first(X1, X2)) -> ACTIVE(first(mark(X1), mark(X2)))
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(s(X)) -> MARK(X)
   MARK(from(X)) -> ACTIVE(from(mark(X)))
   MARK(from(X)) -> MARK(X)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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(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

cons(x1, x2)  =  x1

ACTIVE(x1)  =  x1

mark(x1)  =  x1

from(x1)  =  x1

first(x1, x2)  =  x2

s(x1)  =  s(x1)

active(x1)  =  x1

0  =  0

nil  =  nil


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   s_1=1
   0=2
   nil=1

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

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


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

(37)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   ACTIVE(from(X)) -> MARK(cons(X, from(s(X))))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(first(X1, X2)) -> ACTIVE(first(mark(X1), mark(X2)))
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(from(X)) -> ACTIVE(from(mark(X)))
   MARK(from(X)) -> MARK(X)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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))))
   MARK(from(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

cons(x1, x2)  =  x1

ACTIVE(x1)  =  x1

mark(x1)  =  x1

from(x1)  =  from(x1)

first(x1, x2)  =  x2

s(x1)  =  s

active(x1)  =  x1

0  =  0

nil  =  nil


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   s=2
   0=2
   from_1=1
   nil=1

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

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


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

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

   MARK(cons(X1, X2)) -> ACTIVE(cons(mark(X1), X2))
   MARK(cons(X1, X2)) -> MARK(X1)
   MARK(first(X1, X2)) -> ACTIVE(first(mark(X1), mark(X2)))
   MARK(s(X)) -> ACTIVE(s(mark(X)))
   MARK(from(X)) -> ACTIVE(from(mark(X)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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

(40) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.
----------------------------------------

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

   MARK(cons(X1, X2)) -> MARK(X1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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

(42) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   MARK(cons(X1, X2)) -> MARK(X1)

R is empty.
The set Q consists of the following terms:

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))
   from(mark(x0))
   from(active(x0))

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

(44) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   active(first(0, x0))
   active(first(s(x0), cons(x1, x2)))
   active(from(x0))
   mark(first(x0, x1))
   mark(0)
   mark(nil)
   mark(s(x0))
   mark(cons(x0, x1))
   mark(from(x0))
   first(mark(x0), x1)
   first(x0, mark(x1))
   first(active(x0), x1)
   first(x0, active(x1))
   s(mark(x0))
   s(active(x0))
   from(mark(x0))
   from(active(x0))


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

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

   MARK(cons(X1, X2)) -> MARK(X1)

R is empty.
The set Q consists of the following terms:

   cons(mark(x0), x1)
   cons(x0, mark(x1))
   cons(active(x0), x1)
   cons(x0, active(x1))

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

(46) 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(cons(X1, X2)) -> MARK(X1)
The graph contains the following edges 1 > 1


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(47)
YES