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


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YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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, 21 ms]
(4) QTRS
(5) QTRSRRRProof [EQUIVALENT, 2 ms]
(6) QTRS
(7) QTRSRRRProof [EQUIVALENT, 2 ms]
(8) QTRS
(9) DependencyPairsProof [EQUIVALENT, 0 ms]
(10) QDP
(11) QDPOrderProof [EQUIVALENT, 69 ms]
(12) QDP
(13) DependencyGraphProof [EQUIVALENT, 0 ms]
(14) AND
    (15) QDP
        (16) UsableRulesProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) UsableRulesProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES


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

   a(b(x)) -> b(a(a(x)))
   b(c(x)) -> c(b(b(x)))
   c(a(x)) -> a(c(c(x)))
   u(a(x)) -> x
   v(b(x)) -> x
   w(c(x)) -> x
   a(u(x)) -> x
   b(v(x)) -> x
   c(w(x)) -> x

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(x)) -> a(a(b(x)))
   c(b(x)) -> b(b(c(x)))
   a(c(x)) -> c(c(a(x)))
   a(u(x)) -> x
   b(v(x)) -> x
   c(w(x)) -> x
   u(a(x)) -> x
   v(b(x)) -> x
   w(c(x)) -> x

Q is empty.

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

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

   POL(a(x_1)) = x_1
   POL(b(x_1)) = x_1
   POL(c(x_1)) = x_1
   POL(u(x_1)) = 1 + x_1
   POL(v(x_1)) = x_1
   POL(w(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   a(u(x)) -> x
   u(a(x)) -> x




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

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

   b(a(x)) -> a(a(b(x)))
   c(b(x)) -> b(b(c(x)))
   a(c(x)) -> c(c(a(x)))
   b(v(x)) -> x
   c(w(x)) -> x
   v(b(x)) -> x
   w(c(x)) -> x

Q is empty.

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

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

   POL(a(x_1)) = x_1
   POL(b(x_1)) = x_1
   POL(c(x_1)) = x_1
   POL(v(x_1)) = 1 + x_1
   POL(w(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   b(v(x)) -> x
   v(b(x)) -> x




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

(6)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   b(a(x)) -> a(a(b(x)))
   c(b(x)) -> b(b(c(x)))
   a(c(x)) -> c(c(a(x)))
   c(w(x)) -> x
   w(c(x)) -> x

Q is empty.

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

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

   POL(a(x_1)) = x_1
   POL(b(x_1)) = x_1
   POL(c(x_1)) = x_1
   POL(w(x_1)) = 1 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   c(w(x)) -> x
   w(c(x)) -> x




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

(8)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   b(a(x)) -> a(a(b(x)))
   c(b(x)) -> b(b(c(x)))
   a(c(x)) -> c(c(a(x)))

Q is empty.

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

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

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   b(a(x)) -> a(a(b(x)))
   c(b(x)) -> b(b(c(x)))
   a(c(x)) -> c(c(a(x)))

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

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


The following pairs can be oriented strictly and are deleted.

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

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

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

   a(c(x)) -> c(c(a(x)))
   c(b(x)) -> b(b(c(x)))
   b(a(x)) -> a(a(b(x)))


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

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

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

The TRS R consists of the following rules:

   b(a(x)) -> a(a(b(x)))
   c(b(x)) -> b(b(c(x)))
   a(c(x)) -> c(c(a(x)))

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

(13) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.
----------------------------------------

(14)
Complex Obligation (AND)

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

   C(b(x)) -> C(x)

The TRS R consists of the following rules:

   b(a(x)) -> a(a(b(x)))
   c(b(x)) -> b(b(c(x)))
   a(c(x)) -> c(c(a(x)))

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

(16) 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.
----------------------------------------

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

   C(b(x)) -> C(x)

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

(18) 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(x)) -> C(x)
The graph contains the following edges 1 > 1


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(19)
YES

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

   A(c(x)) -> A(x)

The TRS R consists of the following rules:

   b(a(x)) -> a(a(b(x)))
   c(b(x)) -> b(b(c(x)))
   a(c(x)) -> c(c(a(x)))

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

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

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

   A(c(x)) -> A(x)

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

(23) 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(x)) -> A(x)
The graph contains the following edges 1 > 1


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(24)
YES