YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 28 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) MRRProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) TransformationProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) TransformationProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) UsableRulesProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) TransformationProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) UsableRulesProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) TransformationProof [EQUIVALENT, 0 ms]
        (28) QDP
        (29) UsableRulesProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) TransformationProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) DependencyGraphProof [EQUIVALENT, 0 ms]
        (34) TRUE


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

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

   active(f(b, X, c)) -> mark(f(X, c, X))
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))


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

(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(f(b, X, c)) -> MARK(f(X, c, X))
   ACTIVE(f(b, X, c)) -> F(X, c, X)
   ACTIVE(c) -> MARK(b)
   MARK(f(X1, X2, X3)) -> ACTIVE(f(X1, mark(X2), X3))
   MARK(f(X1, X2, X3)) -> F(X1, mark(X2), X3)
   MARK(f(X1, X2, X3)) -> MARK(X2)
   MARK(b) -> ACTIVE(b)
   MARK(c) -> ACTIVE(c)
   F(mark(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, mark(X2), X3) -> F(X1, X2, X3)
   F(X1, X2, mark(X3)) -> F(X1, X2, X3)
   F(active(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, active(X2), X3) -> F(X1, X2, X3)
   F(X1, X2, active(X3)) -> F(X1, X2, X3)

The TRS R consists of the following rules:

   active(f(b, X, c)) -> mark(f(X, c, X))
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 5 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

   F(X1, mark(X2), X3) -> F(X1, X2, X3)
   F(mark(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, X2, mark(X3)) -> F(X1, X2, X3)
   F(active(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, active(X2), X3) -> F(X1, X2, X3)
   F(X1, X2, active(X3)) -> F(X1, X2, X3)

The TRS R consists of the following rules:

   active(f(b, X, c)) -> mark(f(X, c, X))
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))

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), X3) -> F(X1, X2, X3)
   F(mark(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, X2, mark(X3)) -> F(X1, X2, X3)
   F(active(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, active(X2), X3) -> F(X1, X2, X3)
   F(X1, X2, active(X3)) -> F(X1, X2, X3)

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

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))

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].

   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))


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

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

   F(X1, mark(X2), X3) -> F(X1, X2, X3)
   F(mark(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, X2, mark(X3)) -> F(X1, X2, X3)
   F(active(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, active(X2), X3) -> F(X1, X2, X3)
   F(X1, X2, active(X3)) -> F(X1, X2, X3)

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

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)

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), X3) -> F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3


*F(mark(X1), X2, X3) -> F(X1, X2, X3)
The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3


*F(X1, X2, mark(X3)) -> F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3


*F(active(X1), X2, X3) -> F(X1, X2, X3)
The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3


*F(X1, active(X2), X3) -> F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3


*F(X1, X2, active(X3)) -> F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3


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

(11)
YES

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

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

   MARK(f(X1, X2, X3)) -> ACTIVE(f(X1, mark(X2), X3))
   ACTIVE(f(b, X, c)) -> MARK(f(X, c, X))
   MARK(f(X1, X2, X3)) -> MARK(X2)

The TRS R consists of the following rules:

   active(f(b, X, c)) -> mark(f(X, c, X))
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))

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

(13) 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(f(X1, X2, X3)) -> MARK(X2)


Used ordering: Polynomial interpretation [POLO]:

   POL(ACTIVE(x_1)) = 2*x_1
   POL(MARK(x_1)) = 2*x_1
   POL(active(x_1)) = x_1
   POL(b) = 0
   POL(c) = 0
   POL(f(x_1, x_2, x_3)) = 1 + x_1 + 2*x_2 + x_3
   POL(mark(x_1)) = x_1


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

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MARK(f(X1, X2, X3)) -> ACTIVE(f(X1, mark(X2), X3))
   ACTIVE(f(b, X, c)) -> MARK(f(X, c, X))

The TRS R consists of the following rules:

   active(f(b, X, c)) -> mark(f(X, c, X))
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))

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

(15) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule MARK(f(X1, X2, X3)) -> ACTIVE(f(X1, mark(X2), X3)) we obtained the following new rules [LPAR04]:

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


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

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

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

The TRS R consists of the following rules:

   active(f(b, X, c)) -> mark(f(X, c, X))
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(X1, mark(X2), X3))
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))

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

(17) 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.
----------------------------------------

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   mark(c) -> active(c)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)
   active(c) -> mark(b)
   mark(b) -> active(b)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))

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

(19) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule MARK(f(z0, c, z0)) -> ACTIVE(f(z0, mark(c), z0)) at position [0,1] we obtained the following new rules [LPAR04]:

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


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

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

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

The TRS R consists of the following rules:

   mark(c) -> active(c)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)
   active(c) -> mark(b)
   mark(b) -> active(b)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))

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

(21) 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.
----------------------------------------

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

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

The TRS R consists of the following rules:

   active(c) -> mark(b)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)
   mark(b) -> active(b)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))

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

(23) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule MARK(f(z0, c, z0)) -> ACTIVE(f(z0, active(c), z0)) at position [0,1] we obtained the following new rules [LPAR04]:

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


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

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   active(c) -> mark(b)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)
   mark(b) -> active(b)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))

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

(25) 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.
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(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   mark(b) -> active(b)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))

We have to consider all minimal (P,Q,R)-chains.
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(27) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule MARK(f(z0, c, z0)) -> ACTIVE(f(z0, mark(b), z0)) at position [0,1] we obtained the following new rules [LPAR04]:

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


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

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

The TRS R consists of the following rules:

   mark(b) -> active(b)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))

We have to consider all minimal (P,Q,R)-chains.
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(29) 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.
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(30)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))

We have to consider all minimal (P,Q,R)-chains.
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(31) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule MARK(f(z0, c, z0)) -> ACTIVE(f(z0, active(b), z0)) at position [0] we obtained the following new rules [LPAR04]:

   (MARK(f(x0, c, x0)) -> ACTIVE(f(x0, b, x0)),MARK(f(x0, c, x0)) -> ACTIVE(f(x0, b, x0)))


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

   ACTIVE(f(b, X, c)) -> MARK(f(X, c, X))
   MARK(f(x0, c, x0)) -> ACTIVE(f(x0, b, x0))

The TRS R consists of the following rules:

   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

The set Q consists of the following terms:

   active(f(b, x0, c))
   active(c)
   mark(f(x0, x1, x2))
   mark(b)
   mark(c)
   f(mark(x0), x1, x2)
   f(x0, mark(x1), x2)
   f(x0, x1, mark(x2))
   f(active(x0), x1, x2)
   f(x0, active(x1), x2)
   f(x0, x1, active(x2))

We have to consider all minimal (P,Q,R)-chains.
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(33) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
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(34)
TRUE