/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 36 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 42 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) AND
    (11) QDP
        (12) UsableRulesProof [EQUIVALENT, 1 ms]
        (13) QDP
        (14) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES


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

   a(b(x1)) -> b(c(a(x1)))
   b(c(x1)) -> c(b(b(x1)))
   c(a(x1)) -> a(c(x1))

Q is empty.

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(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:

   b(a(x1)) -> a(c(b(x1)))
   c(b(x1)) -> b(b(c(x1)))
   a(c(x1)) -> c(a(x1))

Q is empty.

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

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

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

   B(a(x1)) -> A(c(b(x1)))
   B(a(x1)) -> C(b(x1))
   B(a(x1)) -> B(x1)
   C(b(x1)) -> B(b(c(x1)))
   C(b(x1)) -> B(c(x1))
   C(b(x1)) -> C(x1)
   A(c(x1)) -> C(a(x1))
   A(c(x1)) -> A(x1)

The TRS R consists of the following rules:

   b(a(x1)) -> a(c(b(x1)))
   c(b(x1)) -> b(b(c(x1)))
   a(c(x1)) -> c(a(x1))

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

(5) 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:

   B(a(x1)) -> C(b(x1))
   B(a(x1)) -> B(x1)


Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = 2 + 2*x_1
   POL(B(x_1)) = 2*x_1
   POL(C(x_1)) = 2*x_1
   POL(a(x_1)) = 1 + x_1
   POL(b(x_1)) = x_1
   POL(c(x_1)) = x_1


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

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

   B(a(x1)) -> A(c(b(x1)))
   C(b(x1)) -> B(b(c(x1)))
   C(b(x1)) -> B(c(x1))
   C(b(x1)) -> C(x1)
   A(c(x1)) -> C(a(x1))
   A(c(x1)) -> A(x1)

The TRS R consists of the following rules:

   b(a(x1)) -> a(c(b(x1)))
   c(b(x1)) -> b(b(c(x1)))
   a(c(x1)) -> c(a(x1))

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

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


The following pairs can be oriented strictly and are deleted.

   B(a(x1)) -> A(c(b(x1)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( A_1(x_1) ) = 2
POL( B_1(x_1) ) = x_1 + 2
POL( C_1(x_1) ) = 2
POL( a_1(x_1) ) = 1
POL( c_1(x_1) ) = 0
POL( b_1(x_1) ) = max{0, 2x_1 - 1}

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

   a(c(x1)) -> c(a(x1))
   c(b(x1)) -> b(b(c(x1)))
   b(a(x1)) -> a(c(b(x1)))


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

(8)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C(b(x1)) -> B(b(c(x1)))
   C(b(x1)) -> B(c(x1))
   C(b(x1)) -> C(x1)
   A(c(x1)) -> C(a(x1))
   A(c(x1)) -> A(x1)

The TRS R consists of the following rules:

   b(a(x1)) -> a(c(b(x1)))
   c(b(x1)) -> b(b(c(x1)))
   a(c(x1)) -> c(a(x1))

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

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

(10)
Complex Obligation (AND)

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

   C(b(x1)) -> C(x1)

The TRS R consists of the following rules:

   b(a(x1)) -> a(c(b(x1)))
   c(b(x1)) -> b(b(c(x1)))
   a(c(x1)) -> c(a(x1))

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

(12) 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.
----------------------------------------

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C(b(x1)) -> C(x1)

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

(14) 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:
*C(b(x1)) -> C(x1)
The graph contains the following edges 1 > 1


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(15)
YES

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

   A(c(x1)) -> A(x1)

The TRS R consists of the following rules:

   b(a(x1)) -> a(c(b(x1)))
   c(b(x1)) -> b(b(c(x1)))
   a(c(x1)) -> c(a(x1))

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

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

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

   A(c(x1)) -> A(x1)

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

(19) 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:
*A(c(x1)) -> A(x1)
The graph contains the following edges 1 > 1


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(20)
YES