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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QDP
(7) QDPOrderProof [EQUIVALENT, 245 ms]
(8) QDP
(9) UsableRulesProof [EQUIVALENT, 0 ms]
(10) QDP
(11) MRRProof [EQUIVALENT, 0 ms]
(12) QDP
(13) PisEmptyProof [EQUIVALENT, 0 ms]
(14) YES


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

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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

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

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

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


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

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


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

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

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

The TRS R consists of the following rules:

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

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

(9) 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.
----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   b(a(x1)) -> x1

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

(11) 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:

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

Strictly oriented rules of the TRS R:

   b(a(x1)) -> x1

Used ordering: Polynomial interpretation [POLO]:

   POL(C(x_1)) = x_1
   POL(a(x_1)) = 3 + x_1
   POL(b(x_1)) = x_1


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

(12)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

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

(14)
YES