YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 1 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 156 ms]
(8) QDP
(9) PisEmptyProof [EQUIVALENT, 0 ms]
(10) YES


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

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

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

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

The TRS R consists of the following rules:

   a(x1) -> x1
   b(a(x1)) -> c(a(b(b(x1))))
   b(b(x1)) -> a(x1)
   c(c(x1)) -> 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 3 less nodes.
----------------------------------------

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

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

The TRS R consists of the following rules:

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

   B(a(x1)) -> B(x1)
   B(a(x1)) -> B(b(x1))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic integers [ARCTIC,STERNAGEL_THIEMANN_RTA14]:

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

   <<<
 POL(a(x_1)) =  	[[2A], [2A], [-1A]] 	 +  	[[2A, -I, 0A], [1A, 0A, -1A], [-I, -I, 0A]] 	* 	x_1
>>>

   <<<
 POL(b(x_1)) =  	[[1A], [-I], [-I]] 	 +  	[[1A, -I, -1A], [2A, -I, 1A], [1A, -1A, -I]] 	* 	x_1
>>>

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


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

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


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

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

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