YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) MNOCProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPOrderProof [EQUIVALENT, 666 ms]
        (13) QDP
        (14) QDPOrderProof [EQUIVALENT, 5205 ms]
        (15) QDP
        (16) PisEmptyProof [EQUIVALENT, 0 ms]
        (17) YES
    (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (22) YES


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

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

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

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

The TRS R consists of the following rules:

   b(b(a(a(x1)))) -> a(a(a(b(b(b(x1))))))
   c(a(x1)) -> a(c(x1))
   c(b(x1)) -> b(c(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 1 less node.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

(10) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   b(b(a(a(x0))))

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

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


The following pairs can be oriented strictly and are deleted.

   B(b(a(a(x1)))) -> B(b(b(x1)))
   B(b(a(a(x1)))) -> B(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

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

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

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


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

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


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

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

   b(b(a(a(x0))))

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

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


The following pairs can be oriented strictly and are deleted.

   B(b(a(a(x1)))) -> B(b(x1))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] to (N^5, +, *, >=, >) :

   <<<
 POL(B(x_1)) =  	[[0]] 	 +  	[[0, 0, 0, 0, 1]] 	* 	x_1
>>>

   <<<
 POL(b(x_1)) =  	[[0], [0], [0], [0], [0]] 	 +  	[[0, 1, 0, 0, 0], [0, 0, 0, 0, 1], [1, 0, 0, 0, 1], [0, 0, 0, 0, 0], [0, 1, 0, 0, 0]] 	* 	x_1
>>>

   <<<
 POL(a(x_1)) =  	[[0], [0], [1], [0], [0]] 	 +  	[[0, 0, 0, 0, 0], [0, 0, 1, 1, 0], [0, 1, 0, 0, 1], [0, 0, 0, 0, 1], [0, 0, 0, 0, 0]] 	* 	x_1
>>>


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

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


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

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

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

The set Q consists of the following terms:

   b(b(a(a(x0))))

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

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

(17)
YES

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

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

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

The TRS R consists of the following rules:

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

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

(19) 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.
----------------------------------------

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

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

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

(21) 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


*C(a(x1)) -> C(x1)
The graph contains the following edges 1 > 1


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

(22)
YES