/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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) DependencyPairsProof [EQUIVALENT, 18 ms]
(2) QDP
(3) QDPOrderProof [EQUIVALENT, 619 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES


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

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

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

The set Q consists of the following terms:

   a(d(x0))
   b(c(x0))
   a(c(x0))
   b(d(x0))


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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   a(d(x0))
   b(c(x0))
   a(c(x0))
   b(d(x0))

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(d(x)) -> B(a(x))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(A(x_1)) =  	[[0A]] 	 +  	[[0A, -I, 0A]] 	* 	x_1
>>>

   <<<
 POL(d(x_1)) =  	[[1A], [0A], [-I]] 	 +  	[[1A, 0A, 0A], [0A, 0A, 0A], [0A, -I, 0A]] 	* 	x_1
>>>

   <<<
 POL(B(x_1)) =  	[[0A]] 	 +  	[[0A, -I, -I]] 	* 	x_1
>>>

   <<<
 POL(a(x_1)) =  	[[0A], [0A], [-I]] 	 +  	[[0A, -I, -I], [0A, 0A, 0A], [0A, -I, -I]] 	* 	x_1
>>>

   <<<
 POL(c(x_1)) =  	[[-I], [0A], [1A]] 	 +  	[[0A, -I, 0A], [0A, 0A, 0A], [0A, 0A, 1A]] 	* 	x_1
>>>

   <<<
 POL(b(x_1)) =  	[[0A], [-I], [-I]] 	 +  	[[-I, -I, 0A], [0A, 0A, 0A], [0A, -I, 0A]] 	* 	x_1
>>>


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

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


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

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

   A(d(x)) -> A(x)
   B(c(x)) -> A(b(x))
   B(c(x)) -> B(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   a(d(x0))
   b(c(x0))
   a(c(x0))
   b(d(x0))

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 1 less node.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   A(d(x)) -> A(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   a(d(x0))
   b(c(x0))
   a(c(x0))
   b(d(x0))

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

(8) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   A(d(x)) -> A(x)

R is empty.
The set Q consists of the following terms:

   a(d(x0))
   b(c(x0))
   a(c(x0))
   b(d(x0))

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

(10) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   a(d(x0))
   b(c(x0))
   a(c(x0))
   b(d(x0))


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

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

   A(d(x)) -> A(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:
*A(d(x)) -> A(x)
The graph contains the following edges 1 > 1


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

(13)
YES

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

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

   B(c(x)) -> B(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   a(d(x0))
   b(c(x0))
   a(c(x0))
   b(d(x0))

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

(15) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

   B(c(x)) -> B(x)

R is empty.
The set Q consists of the following terms:

   a(d(x0))
   b(c(x0))
   a(c(x0))
   b(d(x0))

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

(17) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   a(d(x0))
   b(c(x0))
   a(c(x0))
   b(d(x0))


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

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

   B(c(x)) -> B(x)

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:
*B(c(x)) -> B(x)
The graph contains the following edges 1 > 1


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

(20)
YES