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


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


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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 32 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) QDPOrderProof [EQUIVALENT, 14 ms]
        (17) QDP
        (18) DependencyGraphProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) UsableRulesProof [EQUIVALENT, 0 ms]
        (21) QDP
        (22) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (23) YES


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

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

   active(f(g(X), Y)) -> mark(f(X, f(g(X), Y)))
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(g(X)) -> active(g(mark(X)))
   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)
   g(mark(X)) -> g(X)
   g(active(X)) -> g(X)

Q is empty.

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

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

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

   ACTIVE(f(g(X), Y)) -> MARK(f(X, f(g(X), Y)))
   ACTIVE(f(g(X), Y)) -> F(X, f(g(X), Y))
   MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2))
   MARK(f(X1, X2)) -> F(mark(X1), X2)
   MARK(f(X1, X2)) -> MARK(X1)
   MARK(g(X)) -> ACTIVE(g(mark(X)))
   MARK(g(X)) -> G(mark(X))
   MARK(g(X)) -> MARK(X)
   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)
   G(mark(X)) -> G(X)
   G(active(X)) -> G(X)

The TRS R consists of the following rules:

   active(f(g(X), Y)) -> mark(f(X, f(g(X), Y)))
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(g(X)) -> active(g(mark(X)))
   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)
   g(mark(X)) -> g(X)
   g(active(X)) -> g(X)

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

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

(4)
Complex Obligation (AND)

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

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

   G(active(X)) -> G(X)
   G(mark(X)) -> G(X)

The TRS R consists of the following rules:

   active(f(g(X), Y)) -> mark(f(X, f(g(X), Y)))
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(g(X)) -> active(g(mark(X)))
   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)
   g(mark(X)) -> g(X)
   g(active(X)) -> g(X)

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

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

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

   G(active(X)) -> G(X)
   G(mark(X)) -> G(X)

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

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

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


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


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

(9)
YES

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

(10)
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(f(g(X), Y)) -> mark(f(X, f(g(X), Y)))
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(g(X)) -> active(g(mark(X)))
   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)
   g(mark(X)) -> g(X)
   g(active(X)) -> g(X)

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

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

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

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

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

From the DPs we obtained the following set of size-change graphs:
*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


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

(14)
YES

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

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

   MARK(f(X1, X2)) -> ACTIVE(f(mark(X1), X2))
   ACTIVE(f(g(X), Y)) -> MARK(f(X, f(g(X), Y)))
   MARK(f(X1, X2)) -> MARK(X1)
   MARK(g(X)) -> ACTIVE(g(mark(X)))
   MARK(g(X)) -> MARK(X)

The TRS R consists of the following rules:

   active(f(g(X), Y)) -> mark(f(X, f(g(X), Y)))
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(g(X)) -> active(g(mark(X)))
   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)
   g(mark(X)) -> g(X)
   g(active(X)) -> g(X)

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

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


The following pairs can be oriented strictly and are deleted.

   ACTIVE(f(g(X), Y)) -> MARK(f(X, f(g(X), Y)))
   MARK(g(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

f(x1, x2)  =  x1

ACTIVE(x1)  =  x1

mark(x1)  =  x1

g(x1)  =  g(x1)

active(x1)  =  x1


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   dummyConstant=1
   g_1=1

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

   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   active(f(g(X), Y)) -> mark(f(X, f(g(X), Y)))
   mark(g(X)) -> active(g(mark(X)))
   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)
   g(active(X)) -> g(X)
   g(mark(X)) -> g(X)


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

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

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

The TRS R consists of the following rules:

   active(f(g(X), Y)) -> mark(f(X, f(g(X), Y)))
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(g(X)) -> active(g(mark(X)))
   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)
   g(mark(X)) -> g(X)
   g(active(X)) -> g(X)

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

(18) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(19)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   active(f(g(X), Y)) -> mark(f(X, f(g(X), Y)))
   mark(f(X1, X2)) -> active(f(mark(X1), X2))
   mark(g(X)) -> active(g(mark(X)))
   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)
   g(mark(X)) -> g(X)
   g(active(X)) -> g(X)

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

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

(21)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

(22) 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(f(X1, X2)) -> MARK(X1)
The graph contains the following edges 1 > 1


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

(23)
YES