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


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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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 27 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 19 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) UsableRulesProof [EQUIVALENT, 1 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 1 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES
    (19) QDP
        (20) QDPOrderProof [EQUIVALENT, 25 ms]
        (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (25) YES


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

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

   f(f(X)) -> c(n__f(n__g(n__f(X))))
   c(X) -> d(activate(X))
   h(X) -> c(n__d(X))
   f(X) -> n__f(X)
   g(X) -> n__g(X)
   d(X) -> n__d(X)
   activate(n__f(X)) -> f(activate(X))
   activate(n__g(X)) -> g(X)
   activate(n__d(X)) -> d(X)
   activate(X) -> X

Q is empty.

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

(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(f(X)) -> n__f(n__g(n__f(c(X))))
   c(X) -> activate(d(X))
   h(X) -> n__d(c(X))
   f(X) -> n__f(X)
   g(X) -> n__g(X)
   d(X) -> n__d(X)
   n__f(activate(X)) -> activate(f(X))
   n__g(activate(X)) -> g(X)
   n__d(activate(X)) -> d(X)
   activate(X) -> X

Q is empty.

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

(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(activate(x_1)) = x_1
   POL(c(x_1)) = x_1
   POL(d(x_1)) = x_1
   POL(f(x_1)) = x_1
   POL(g(x_1)) = x_1
   POL(h(x_1)) = 1 + x_1
   POL(n__d(x_1)) = x_1
   POL(n__f(x_1)) = x_1
   POL(n__g(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   h(X) -> n__d(c(X))




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

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(f(X)) -> n__f(n__g(n__f(c(X))))
   c(X) -> activate(d(X))
   f(X) -> n__f(X)
   g(X) -> n__g(X)
   d(X) -> n__d(X)
   n__f(activate(X)) -> activate(f(X))
   n__g(activate(X)) -> g(X)
   n__d(activate(X)) -> d(X)
   activate(X) -> X

Q is empty.

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

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

(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(f(X)) -> N__F(n__g(n__f(c(X))))
   F(f(X)) -> N__G(n__f(c(X)))
   F(f(X)) -> N__F(c(X))
   F(f(X)) -> C(X)
   C(X) -> ACTIVATE(d(X))
   C(X) -> D(X)
   F(X) -> N__F(X)
   G(X) -> N__G(X)
   D(X) -> N__D(X)
   N__F(activate(X)) -> ACTIVATE(f(X))
   N__F(activate(X)) -> F(X)
   N__G(activate(X)) -> G(X)
   N__D(activate(X)) -> D(X)

The TRS R consists of the following rules:

   f(f(X)) -> n__f(n__g(n__f(c(X))))
   c(X) -> activate(d(X))
   f(X) -> n__f(X)
   g(X) -> n__g(X)
   d(X) -> n__d(X)
   n__f(activate(X)) -> activate(f(X))
   n__g(activate(X)) -> g(X)
   n__d(activate(X)) -> d(X)
   activate(X) -> X

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

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

(8)
Complex Obligation (AND)

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

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

   N__D(activate(X)) -> D(X)
   D(X) -> N__D(X)

The TRS R consists of the following rules:

   f(f(X)) -> n__f(n__g(n__f(c(X))))
   c(X) -> activate(d(X))
   f(X) -> n__f(X)
   g(X) -> n__g(X)
   d(X) -> n__d(X)
   n__f(activate(X)) -> activate(f(X))
   n__g(activate(X)) -> g(X)
   n__d(activate(X)) -> d(X)
   activate(X) -> X

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

(10) 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.
----------------------------------------

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   N__D(activate(X)) -> D(X)
   D(X) -> N__D(X)

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

(12) 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:
*D(X) -> N__D(X)
The graph contains the following edges 1 >= 1


*N__D(activate(X)) -> D(X)
The graph contains the following edges 1 > 1


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

(13)
YES

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

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

   N__G(activate(X)) -> G(X)
   G(X) -> N__G(X)

The TRS R consists of the following rules:

   f(f(X)) -> n__f(n__g(n__f(c(X))))
   c(X) -> activate(d(X))
   f(X) -> n__f(X)
   g(X) -> n__g(X)
   d(X) -> n__d(X)
   n__f(activate(X)) -> activate(f(X))
   n__g(activate(X)) -> g(X)
   n__d(activate(X)) -> d(X)
   activate(X) -> X

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

(15) 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.
----------------------------------------

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

   N__G(activate(X)) -> G(X)
   G(X) -> N__G(X)

R is empty.
Q is empty.
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(X) -> N__G(X)
The graph contains the following edges 1 >= 1


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


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

(18)
YES

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

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

   N__F(activate(X)) -> F(X)
   F(f(X)) -> N__F(n__g(n__f(c(X))))
   F(f(X)) -> N__F(c(X))
   F(X) -> N__F(X)

The TRS R consists of the following rules:

   f(f(X)) -> n__f(n__g(n__f(c(X))))
   c(X) -> activate(d(X))
   f(X) -> n__f(X)
   g(X) -> n__g(X)
   d(X) -> n__d(X)
   n__f(activate(X)) -> activate(f(X))
   n__g(activate(X)) -> g(X)
   n__d(activate(X)) -> d(X)
   activate(X) -> X

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

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


The following pairs can be oriented strictly and are deleted.

   F(f(X)) -> N__F(n__g(n__f(c(X))))
   F(f(X)) -> N__F(c(X))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(F(x_1)) = x_1
   POL(N__F(x_1)) = x_1
   POL(activate(x_1)) = x_1
   POL(c(x_1)) = x_1
   POL(d(x_1)) = x_1
   POL(f(x_1)) = 1 + x_1
   POL(g(x_1)) = 0
   POL(n__d(x_1)) = x_1
   POL(n__f(x_1)) = 0
   POL(n__g(x_1)) = 0

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

   c(X) -> activate(d(X))
   n__g(activate(X)) -> g(X)
   g(X) -> n__g(X)
   activate(X) -> X
   n__d(activate(X)) -> d(X)
   d(X) -> n__d(X)


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

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

   N__F(activate(X)) -> F(X)
   F(X) -> N__F(X)

The TRS R consists of the following rules:

   f(f(X)) -> n__f(n__g(n__f(c(X))))
   c(X) -> activate(d(X))
   f(X) -> n__f(X)
   g(X) -> n__g(X)
   d(X) -> n__d(X)
   n__f(activate(X)) -> activate(f(X))
   n__g(activate(X)) -> g(X)
   n__d(activate(X)) -> d(X)
   activate(X) -> X

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

(22) 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.
----------------------------------------

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

   N__F(activate(X)) -> F(X)
   F(X) -> N__F(X)

R is empty.
Q is empty.
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:
*F(X) -> N__F(X)
The graph contains the following edges 1 >= 1


*N__F(activate(X)) -> F(X)
The graph contains the following edges 1 > 1


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

(25)
YES