/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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, 23 ms]
(4) QTRS
(5) QTRSRRRProof [EQUIVALENT, 3 ms]
(6) QTRS
(7) QTRSRRRProof [EQUIVALENT, 2 ms]
(8) QTRS
(9) DependencyPairsProof [EQUIVALENT, 19 ms]
(10) QDP
(11) MRRProof [EQUIVALENT, 100 ms]
(12) QDP
(13) DependencyGraphProof [EQUIVALENT, 3 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
    (25) QDP
        (26) MRRProof [EQUIVALENT, 12 ms]
        (27) QDP
        (28) PisEmptyProof [EQUIVALENT, 0 ms]
        (29) YES


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

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

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

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:

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

Q is empty.

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

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

   POL(A(x_1)) = 1 + x_1
   POL(B(x_1)) = x_1
   POL(C(x_1)) = x_1
   POL(a(x_1)) = x_1
   POL(b(x_1)) = x_1
   POL(c(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(a(x1)) -> x1
   a(A(x1)) -> x1




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

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

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

Q is empty.

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

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

   POL(A(x_1)) = x_1
   POL(B(x_1)) = 1 + x_1
   POL(C(x_1)) = x_1
   POL(a(x_1)) = x_1
   POL(b(x_1)) = x_1
   POL(c(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(b(x1)) -> x1
   b(B(x1)) -> x1




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

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

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

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

   C(c(x1)) -> x1
   c(C(x1)) -> x1




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

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

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

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:

   C^1(b(a(x1))) -> A^1(b(c(x1)))
   C^1(b(a(x1))) -> B^1(c(x1))
   C^1(b(a(x1))) -> C^1(x1)
   A^2(B(C(x1))) -> C^2(B(A(x1)))
   A^2(B(C(x1))) -> B^2(A(x1))
   A^2(B(C(x1))) -> A^2(x1)
   C^2(a(b(x1))) -> B^1(a(C(x1)))
   C^2(a(b(x1))) -> A^1(C(x1))
   C^2(a(b(x1))) -> C^2(x1)
   B^2(A(c(x1))) -> C^1(A(B(x1)))
   B^2(A(c(x1))) -> A^2(B(x1))
   B^2(A(c(x1))) -> B^2(x1)
   B^1(c(A(x1))) -> A^2(c(b(x1)))
   B^1(c(A(x1))) -> C^1(b(x1))
   B^1(c(A(x1))) -> B^1(x1)
   A^1(C(B(x1))) -> B^2(C(a(x1)))
   A^1(C(B(x1))) -> C^2(a(x1))
   A^1(C(B(x1))) -> A^1(x1)

The TRS R consists of the following rules:

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

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

(11) 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:

   C^1(b(a(x1))) -> A^1(b(c(x1)))
   C^1(b(a(x1))) -> C^1(x1)
   A^2(B(C(x1))) -> C^2(B(A(x1)))
   A^2(B(C(x1))) -> B^2(A(x1))
   C^2(a(b(x1))) -> B^1(a(C(x1)))
   C^2(a(b(x1))) -> A^1(C(x1))
   C^2(a(b(x1))) -> C^2(x1)
   B^2(A(c(x1))) -> C^1(A(B(x1)))
   B^2(A(c(x1))) -> A^2(B(x1))
   B^2(A(c(x1))) -> B^2(x1)
   B^1(c(A(x1))) -> B^1(x1)

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = x_1
   POL(A^1(x_1)) = x_1
   POL(A^2(x_1)) = 1 + x_1
   POL(B(x_1)) = 2*x_1
   POL(B^1(x_1)) = 3 + 2*x_1
   POL(B^2(x_1)) = 2*x_1
   POL(C(x_1)) = 2*x_1
   POL(C^1(x_1)) = 3 + 2*x_1
   POL(C^2(x_1)) = 2*x_1
   POL(a(x_1)) = x_1
   POL(b(x_1)) = 2 + 2*x_1
   POL(c(x_1)) = 2 + 2*x_1


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

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

   C^1(b(a(x1))) -> B^1(c(x1))
   A^2(B(C(x1))) -> A^2(x1)
   B^1(c(A(x1))) -> A^2(c(b(x1)))
   B^1(c(A(x1))) -> C^1(b(x1))
   A^1(C(B(x1))) -> B^2(C(a(x1)))
   A^1(C(B(x1))) -> C^2(a(x1))
   A^1(C(B(x1))) -> A^1(x1)

The TRS R consists of the following rules:

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

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 3 SCCs with 3 less nodes.
----------------------------------------

(14)
Complex Obligation (AND)

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

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

   A^1(C(B(x1))) -> A^1(x1)

The TRS R consists of the following rules:

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

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:

   A^1(C(B(x1))) -> A^1(x1)

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:
*A^1(C(B(x1))) -> A^1(x1)
The graph contains the following edges 1 > 1


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

(19)
YES

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

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

   A^2(B(C(x1))) -> A^2(x1)

The TRS R consists of the following rules:

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

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^2(B(C(x1))) -> A^2(x1)

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^2(B(C(x1))) -> A^2(x1)
The graph contains the following edges 1 > 1


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

(24)
YES

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

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B^1(c(A(x1))) -> C^1(b(x1))
   C^1(b(a(x1))) -> B^1(c(x1))

The TRS R consists of the following rules:

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

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

(26) 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^1(c(A(x1))) -> C^1(b(x1))
   C^1(b(a(x1))) -> B^1(c(x1))

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = 1 + x_1
   POL(B(x_1)) = x_1
   POL(B^1(x_1)) = 3 + 2*x_1
   POL(C(x_1)) = x_1
   POL(C^1(x_1)) = 2 + 2*x_1
   POL(a(x_1)) = x_1
   POL(b(x_1)) = 3 + 2*x_1
   POL(c(x_1)) = 2 + 2*x_1


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

(27)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

(28) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(29)
YES