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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) QDPOrderProof [EQUIVALENT, 54 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 336 ms]
(8) QDP
(9) PisEmptyProof [EQUIVALENT, 0 ms]
(10) YES


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

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

   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
   1(1(0(1(x1)))) -> 0(0(0(0(x1))))

Q is empty.

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

(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:

   0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1))))
   0^1(0(0(0(x1)))) -> 1^1(1(1(x1)))
   0^1(0(0(0(x1)))) -> 1^1(1(x1))
   0^1(0(0(0(x1)))) -> 1^1(x1)
   1^1(1(0(1(x1)))) -> 0^1(0(0(0(x1))))
   1^1(1(0(1(x1)))) -> 0^1(0(0(x1)))
   1^1(1(0(1(x1)))) -> 0^1(0(x1))
   1^1(1(0(1(x1)))) -> 0^1(x1)

The TRS R consists of the following rules:

   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
   1(1(0(1(x1)))) -> 0(0(0(0(x1))))

Q is empty.
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.

   0^1(0(0(0(x1)))) -> 1^1(1(1(x1)))
   0^1(0(0(0(x1)))) -> 1^1(1(x1))
   0^1(0(0(0(x1)))) -> 1^1(x1)
   1^1(1(0(1(x1)))) -> 0^1(0(0(x1)))
   1^1(1(0(1(x1)))) -> 0^1(0(x1))
   1^1(1(0(1(x1)))) -> 0^1(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(0(x_1)) = 1 + x_1
   POL(0^1(x_1)) = 1 + x_1
   POL(1(x_1)) = 1 + x_1
   POL(1^1(x_1)) = 1 + x_1

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

   1(1(0(1(x1)))) -> 0(0(0(0(x1))))
   0(0(0(0(x1)))) -> 0(1(1(1(x1))))


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

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

   0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1))))
   1^1(1(0(1(x1)))) -> 0^1(0(0(0(x1))))

The TRS R consists of the following rules:

   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
   1(1(0(1(x1)))) -> 0(0(0(0(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 1 SCC with 1 less node.
----------------------------------------

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

   0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1))))

The TRS R consists of the following rules:

   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
   1(1(0(1(x1)))) -> 0(0(0(0(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.

   0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

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

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

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


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

   1(1(0(1(x1)))) -> 0(0(0(0(x1))))
   0(0(0(0(x1)))) -> 0(1(1(1(x1))))


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

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

   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
   1(1(0(1(x1)))) -> 0(0(0(0(x1))))

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

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

(10)
YES