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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 16 ms]
(2) QDP
(3) QDPOrderProof [EQUIVALENT, 26 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (16) YES


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

   a(b(x1)) -> b(r(x1))
   r(a(x1)) -> d(r(x1))
   r(x1) -> d(x1)
   d(a(x1)) -> a(a(d(x1)))
   d(x1) -> a(x1)

Q is empty.

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(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:

   A(b(x1)) -> R(x1)
   R(a(x1)) -> D(r(x1))
   R(a(x1)) -> R(x1)
   R(x1) -> D(x1)
   D(a(x1)) -> A(a(d(x1)))
   D(a(x1)) -> A(d(x1))
   D(a(x1)) -> D(x1)
   D(x1) -> A(x1)

The TRS R consists of the following rules:

   a(b(x1)) -> b(r(x1))
   r(a(x1)) -> d(r(x1))
   r(x1) -> d(x1)
   d(a(x1)) -> a(a(d(x1)))
   d(x1) -> a(x1)

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

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


The following pairs can be oriented strictly and are deleted.

   A(b(x1)) -> R(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(A(x_1)) = x_1
   POL(D(x_1)) = x_1
   POL(R(x_1)) = x_1
   POL(a(x_1)) = x_1
   POL(b(x_1)) = 1 + x_1
   POL(d(x_1)) = x_1
   POL(r(x_1)) = x_1

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

   r(a(x1)) -> d(r(x1))
   r(x1) -> d(x1)
   d(a(x1)) -> a(a(d(x1)))
   d(x1) -> a(x1)
   a(b(x1)) -> b(r(x1))


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

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   R(a(x1)) -> D(r(x1))
   R(a(x1)) -> R(x1)
   R(x1) -> D(x1)
   D(a(x1)) -> A(a(d(x1)))
   D(a(x1)) -> A(d(x1))
   D(a(x1)) -> D(x1)
   D(x1) -> A(x1)

The TRS R consists of the following rules:

   a(b(x1)) -> b(r(x1))
   r(a(x1)) -> d(r(x1))
   r(x1) -> d(x1)
   d(a(x1)) -> a(a(d(x1)))
   d(x1) -> a(x1)

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

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

(6)
Complex Obligation (AND)

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

   D(a(x1)) -> D(x1)

The TRS R consists of the following rules:

   a(b(x1)) -> b(r(x1))
   r(a(x1)) -> d(r(x1))
   r(x1) -> d(x1)
   d(a(x1)) -> a(a(d(x1)))
   d(x1) -> a(x1)

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

(8) 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.
----------------------------------------

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

   D(a(x1)) -> D(x1)

R is empty.
Q is empty.
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:
*D(a(x1)) -> D(x1)
The graph contains the following edges 1 > 1


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

(11)
YES

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

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

   R(a(x1)) -> R(x1)

The TRS R consists of the following rules:

   a(b(x1)) -> b(r(x1))
   r(a(x1)) -> d(r(x1))
   r(x1) -> d(x1)
   d(a(x1)) -> a(a(d(x1)))
   d(x1) -> a(x1)

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

(13) 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.
----------------------------------------

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

   R(a(x1)) -> R(x1)

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

(15) 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:
*R(a(x1)) -> R(x1)
The graph contains the following edges 1 > 1


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

(16)
YES