/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 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) MRRProof [EQUIVALENT, 27 ms]
        (28) QDP
        (29) TransformationProof [EQUIVALENT, 2 ms]
        (30) QDP
        (31) TransformationProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) TransformationProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) TransformationProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) TransformationProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) TransformationProof [EQUIVALENT, 0 ms]
        (40) QDP
        (41) TransformationProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) TransformationProof [EQUIVALENT, 0 ms]
        (44) QDP
        (45) QDPOrderProof [EQUIVALENT, 106 ms]
        (46) QDP
        (47) DependencyGraphProof [EQUIVALENT, 0 ms]
        (48) QDP
        (49) QDPOrderProof [EQUIVALENT, 0 ms]
        (50) QDP
        (51) PisEmptyProof [EQUIVALENT, 0 ms]
        (52) YES


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

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

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))


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

(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(h(X)) -> MARK(g(X, X))
   ACTIVE(h(X)) -> G(X, X)
   ACTIVE(g(a, X)) -> MARK(f(b, X))
   ACTIVE(g(a, X)) -> F(b, X)
   ACTIVE(f(X, X)) -> MARK(h(a))
   ACTIVE(f(X, X)) -> H(a)
   ACTIVE(a) -> MARK(b)
   MARK(h(X)) -> ACTIVE(h(mark(X)))
   MARK(h(X)) -> H(mark(X))
   MARK(h(X)) -> MARK(X)
   MARK(g(X1, X2)) -> ACTIVE(g(mark(X1), X2))
   MARK(g(X1, X2)) -> G(mark(X1), X2)
   MARK(g(X1, X2)) -> MARK(X1)
   MARK(a) -> ACTIVE(a)
   MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2))
   MARK(f(X1, X2)) -> F(mark(X1), X2)
   MARK(f(X1, X2)) -> MARK(X1)
   MARK(b) -> ACTIVE(b)
   H(mark(X)) -> H(X)
   H(active(X)) -> H(X)
   G(mark(X1), X2) -> G(X1, X2)
   G(X1, mark(X2)) -> G(X1, X2)
   G(active(X1), X2) -> G(X1, X2)
   G(X1, active(X2)) -> G(X1, X2)
   F(mark(X1), X2) -> F(X1, X2)
   F(X1, mark(X2)) -> F(X1, X2)
   F(active(X1), X2) -> F(X1, X2)
   F(X1, active(X2)) -> F(X1, X2)

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 9 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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:

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

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

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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].

   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))


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

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)

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


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


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


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


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

(11)
YES

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

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

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

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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:

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

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

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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].

   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))


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

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)

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


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


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


*G(X1, active(X2)) -> G(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:

   H(active(X)) -> H(X)
   H(mark(X)) -> H(X)

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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:

   H(active(X)) -> H(X)
   H(mark(X)) -> H(X)

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

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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].

   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))


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

(23)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   H(active(X)) -> H(X)
   H(mark(X)) -> H(X)

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

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)

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:
*H(active(X)) -> H(X)
The graph contains the following edges 1 > 1


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


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

(25)
YES

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

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

   MARK(g(X1, X2)) -> ACTIVE(g(mark(X1), X2))
   ACTIVE(h(X)) -> MARK(g(X, X))
   MARK(g(X1, X2)) -> MARK(X1)
   MARK(h(X)) -> ACTIVE(h(mark(X)))
   ACTIVE(g(a, X)) -> MARK(f(b, X))
   MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2))
   ACTIVE(f(X, X)) -> MARK(h(a))
   MARK(h(X)) -> MARK(X)
   MARK(f(X1, X2)) -> MARK(X1)

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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

(27) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   MARK(g(X1, X2)) -> MARK(X1)
   MARK(h(X)) -> MARK(X)
   MARK(f(X1, X2)) -> MARK(X1)


Used ordering: Polynomial interpretation [POLO]:

   POL(ACTIVE(x_1)) = 2*x_1
   POL(MARK(x_1)) = 2*x_1
   POL(a) = 0
   POL(active(x_1)) = x_1
   POL(b) = 0
   POL(f(x_1, x_2)) = 2 + 2*x_1 + x_2
   POL(g(x_1, x_2)) = 2 + x_1 + x_2
   POL(h(x_1)) = 2 + 2*x_1
   POL(mark(x_1)) = x_1


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

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

   MARK(g(X1, X2)) -> ACTIVE(g(mark(X1), X2))
   ACTIVE(h(X)) -> MARK(g(X, X))
   MARK(h(X)) -> ACTIVE(h(mark(X)))
   ACTIVE(g(a, X)) -> MARK(f(b, X))
   MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2))
   ACTIVE(f(X, X)) -> MARK(h(a))

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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

(29) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule MARK(g(X1, X2)) -> ACTIVE(g(mark(X1), X2)) we obtained the following new rules [LPAR04]:

   (MARK(g(z0, z0)) -> ACTIVE(g(mark(z0), z0)),MARK(g(z0, z0)) -> ACTIVE(g(mark(z0), z0)))


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

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

   ACTIVE(h(X)) -> MARK(g(X, X))
   MARK(h(X)) -> ACTIVE(h(mark(X)))
   ACTIVE(g(a, X)) -> MARK(f(b, X))
   MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2))
   ACTIVE(f(X, X)) -> MARK(h(a))
   MARK(g(z0, z0)) -> ACTIVE(g(mark(z0), z0))

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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

(31) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule MARK(h(X)) -> ACTIVE(h(mark(X))) we obtained the following new rules [LPAR04]:

   (MARK(h(a)) -> ACTIVE(h(mark(a))),MARK(h(a)) -> ACTIVE(h(mark(a))))


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

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

   ACTIVE(h(X)) -> MARK(g(X, X))
   ACTIVE(g(a, X)) -> MARK(f(b, X))
   MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2))
   ACTIVE(f(X, X)) -> MARK(h(a))
   MARK(g(z0, z0)) -> ACTIVE(g(mark(z0), z0))
   MARK(h(a)) -> ACTIVE(h(mark(a)))

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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

(33) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule MARK(h(a)) -> ACTIVE(h(mark(a))) at position [0,0] we obtained the following new rules [LPAR04]:

   (MARK(h(a)) -> ACTIVE(h(active(a))),MARK(h(a)) -> ACTIVE(h(active(a))))


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

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

   ACTIVE(h(X)) -> MARK(g(X, X))
   ACTIVE(g(a, X)) -> MARK(f(b, X))
   MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2))
   ACTIVE(f(X, X)) -> MARK(h(a))
   MARK(g(z0, z0)) -> ACTIVE(g(mark(z0), z0))
   MARK(h(a)) -> ACTIVE(h(active(a)))

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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

(35) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule MARK(h(a)) -> ACTIVE(h(active(a))) at position [0,0] we obtained the following new rules [LPAR04]:

   (MARK(h(a)) -> ACTIVE(h(mark(b))),MARK(h(a)) -> ACTIVE(h(mark(b))))


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

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

   ACTIVE(h(X)) -> MARK(g(X, X))
   ACTIVE(g(a, X)) -> MARK(f(b, X))
   MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2))
   ACTIVE(f(X, X)) -> MARK(h(a))
   MARK(g(z0, z0)) -> ACTIVE(g(mark(z0), z0))
   MARK(h(a)) -> ACTIVE(h(mark(b)))

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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

(37) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule MARK(h(a)) -> ACTIVE(h(mark(b))) at position [0,0] we obtained the following new rules [LPAR04]:

   (MARK(h(a)) -> ACTIVE(h(active(b))),MARK(h(a)) -> ACTIVE(h(active(b))))


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

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

   ACTIVE(h(X)) -> MARK(g(X, X))
   ACTIVE(g(a, X)) -> MARK(f(b, X))
   MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2))
   ACTIVE(f(X, X)) -> MARK(h(a))
   MARK(g(z0, z0)) -> ACTIVE(g(mark(z0), z0))
   MARK(h(a)) -> ACTIVE(h(active(b)))

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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

(39) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule MARK(h(a)) -> ACTIVE(h(active(b))) at position [0] we obtained the following new rules [LPAR04]:

   (MARK(h(a)) -> ACTIVE(h(b)),MARK(h(a)) -> ACTIVE(h(b)))


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

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

   ACTIVE(h(X)) -> MARK(g(X, X))
   ACTIVE(g(a, X)) -> MARK(f(b, X))
   MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2))
   ACTIVE(f(X, X)) -> MARK(h(a))
   MARK(g(z0, z0)) -> ACTIVE(g(mark(z0), z0))
   MARK(h(a)) -> ACTIVE(h(b))

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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

(41) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2)) we obtained the following new rules [LPAR04]:

   (MARK(f(b, z0)) -> ACTIVE(f(mark(b), z0)),MARK(f(b, z0)) -> ACTIVE(f(mark(b), z0)))


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

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

   ACTIVE(h(X)) -> MARK(g(X, X))
   ACTIVE(g(a, X)) -> MARK(f(b, X))
   ACTIVE(f(X, X)) -> MARK(h(a))
   MARK(g(z0, z0)) -> ACTIVE(g(mark(z0), z0))
   MARK(h(a)) -> ACTIVE(h(b))
   MARK(f(b, z0)) -> ACTIVE(f(mark(b), z0))

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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

(43) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule MARK(f(b, z0)) -> ACTIVE(f(mark(b), z0)) at position [0,0] we obtained the following new rules [LPAR04]:

   (MARK(f(b, z0)) -> ACTIVE(f(active(b), z0)),MARK(f(b, z0)) -> ACTIVE(f(active(b), z0)))


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

(44)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(h(X)) -> MARK(g(X, X))
   ACTIVE(g(a, X)) -> MARK(f(b, X))
   ACTIVE(f(X, X)) -> MARK(h(a))
   MARK(g(z0, z0)) -> ACTIVE(g(mark(z0), z0))
   MARK(h(a)) -> ACTIVE(h(b))
   MARK(f(b, z0)) -> ACTIVE(f(active(b), z0))

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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

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


The following pairs can be oriented strictly and are deleted.

   MARK(h(a)) -> ACTIVE(h(b))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( ACTIVE_1(x_1) ) = max{0, x_1 - 1}
POL( g_2(x_1, x_2) ) = x_1 + 1
POL( mark_1(x_1) ) = x_1
POL( active_1(x_1) ) = x_1
POL( h_1(x_1) ) = x_1 + 1
POL( a ) = 1
POL( f_2(x_1, x_2) ) = 2
POL( b ) = 0
POL( MARK_1(x_1) ) = max{0, x_1 - 1}

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

   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   active(f(X, X)) -> mark(h(a))
   mark(h(X)) -> active(h(mark(X)))
   mark(a) -> active(a)
   mark(b) -> active(b)
   g(X1, mark(X2)) -> g(X1, X2)
   g(mark(X1), X2) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)
   active(a) -> mark(b)
   h(active(X)) -> h(X)
   h(mark(X)) -> h(X)


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

(46)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(h(X)) -> MARK(g(X, X))
   ACTIVE(g(a, X)) -> MARK(f(b, X))
   ACTIVE(f(X, X)) -> MARK(h(a))
   MARK(g(z0, z0)) -> ACTIVE(g(mark(z0), z0))
   MARK(f(b, z0)) -> ACTIVE(f(active(b), z0))

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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

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

(48)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(g(z0, z0)) -> ACTIVE(g(mark(z0), z0))
   ACTIVE(h(X)) -> MARK(g(X, X))

The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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

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


The following pairs can be oriented strictly and are deleted.

   MARK(g(z0, z0)) -> ACTIVE(g(mark(z0), z0))
   ACTIVE(h(X)) -> MARK(g(X, X))
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
MARK(x1)  =  MARK

ACTIVE(x1)  =  x1

g(x1, x2)  =  g

h(x1)  =  h


Knuth-Bendix order [KBO] with precedence:MARK > g

and weight map:

   MARK=1
   h=2
   g=1

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

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


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

(50)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   active(h(X)) -> mark(g(X, X))
   active(g(a, X)) -> mark(f(b, X))
   active(f(X, X)) -> mark(h(a))
   active(a) -> mark(b)
   mark(h(X)) -> active(h(mark(X)))
   mark(g(X1, X2)) -> active(g(mark(X1), X2))
   mark(a) -> active(a)
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(b) -> active(b)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
   g(mark(X1), X2) -> g(X1, X2)
   g(X1, mark(X2)) -> g(X1, X2)
   g(active(X1), X2) -> g(X1, X2)
   g(X1, active(X2)) -> g(X1, X2)
   f(mark(X1), X2) -> f(X1, X2)
   f(X1, mark(X2)) -> f(X1, X2)
   f(active(X1), X2) -> f(X1, X2)
   f(X1, active(X2)) -> f(X1, X2)

The set Q consists of the following terms:

   active(h(x0))
   active(g(a, x0))
   active(f(x0, x0))
   active(a)
   mark(h(x0))
   mark(g(x0, x1))
   mark(a)
   mark(f(x0, x1))
   mark(b)
   h(mark(x0))
   h(active(x0))
   g(mark(x0), x1)
   g(x0, mark(x1))
   g(active(x0), x1)
   g(x0, active(x1))
   f(mark(x0), x1)
   f(x0, mark(x1))
   f(active(x0), x1)
   f(x0, active(x1))

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

(51) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(52)
YES