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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRSRRRProof [EQUIVALENT, 71 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 18 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 111 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 366 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) TRUE


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

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

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

Q is empty.

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

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

   POL(a(x_1)) = 1 + x_1
   POL(b(x_1)) = 3 + x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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




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

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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


Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = x_1
   POL(B(x_1)) = 2 + x_1
   POL(a(x_1)) = 1 + x_1
   POL(b(x_1)) = 3 + x_1


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

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

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

The TRS R consists of the following rules:

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

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

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

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

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

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


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

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


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

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

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

The TRS R consists of the following rules:

   a(a(a(b(b(x1))))) -> b(b(b(x1)))
   b(a(a(a(b(x1))))) -> a(a(a(b(a(a(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 0 SCCs with 1 less node.
----------------------------------------

(10)
TRUE