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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 16 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 92 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 30 ms]
(8) QDP
(9) PisEmptyProof [EQUIVALENT, 0 ms]
(10) YES


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

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

   c(c(c(a(x, y)))) -> b(c(c(c(c(y)))), x)
   c(c(b(c(y), 0))) -> a(0, c(c(a(y, 0))))
   c(c(a(a(y, 0), x))) -> c(y)

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:

   C(c(c(a(x, y)))) -> C(c(c(c(y))))
   C(c(c(a(x, y)))) -> C(c(c(y)))
   C(c(c(a(x, y)))) -> C(c(y))
   C(c(c(a(x, y)))) -> C(y)
   C(c(b(c(y), 0))) -> C(c(a(y, 0)))
   C(c(b(c(y), 0))) -> C(a(y, 0))
   C(c(a(a(y, 0), x))) -> C(y)

The TRS R consists of the following rules:

   c(c(c(a(x, y)))) -> b(c(c(c(c(y)))), x)
   c(c(b(c(y), 0))) -> a(0, c(c(a(y, 0))))
   c(c(a(a(y, 0), x))) -> c(y)

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
----------------------------------------

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

   C(c(c(a(x, y)))) -> C(c(c(y)))
   C(c(c(a(x, y)))) -> C(c(c(c(y))))
   C(c(c(a(x, y)))) -> C(c(y))
   C(c(c(a(x, y)))) -> C(y)
   C(c(b(c(y), 0))) -> C(c(a(y, 0)))
   C(c(a(a(y, 0), x))) -> C(y)

The TRS R consists of the following rules:

   c(c(c(a(x, y)))) -> b(c(c(c(c(y)))), x)
   c(c(b(c(y), 0))) -> a(0, c(c(a(y, 0))))
   c(c(a(a(y, 0), x))) -> c(y)

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

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


The following pairs can be oriented strictly and are deleted.

   C(c(c(a(x, y)))) -> C(c(c(y)))
   C(c(c(a(x, y)))) -> C(c(y))
   C(c(c(a(x, y)))) -> C(y)
   C(c(b(c(y), 0))) -> C(c(a(y, 0)))
   C(c(a(a(y, 0), x))) -> C(y)
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

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

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

   <<<
 POL(a(x_1, x_2)) =  	[[0A]] 	 +  	[[0A]] 	* 	x_1 	 +  	[[1A]] 	* 	x_2
>>>

   <<<
 POL(b(x_1, x_2)) =  	[[2A]] 	 +  	[[0A]] 	* 	x_1 	 +  	[[0A]] 	* 	x_2
>>>

   <<<
 POL(0) =  	[[0A]]
>>>


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

   c(c(c(a(x, y)))) -> b(c(c(c(c(y)))), x)
   c(c(b(c(y), 0))) -> a(0, c(c(a(y, 0))))
   c(c(a(a(y, 0), x))) -> c(y)


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

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

   C(c(c(a(x, y)))) -> C(c(c(c(y))))

The TRS R consists of the following rules:

   c(c(c(a(x, y)))) -> b(c(c(c(c(y)))), x)
   c(c(b(c(y), 0))) -> a(0, c(c(a(y, 0))))
   c(c(a(a(y, 0), x))) -> c(y)

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

   <<<
 POL(C(x_1)) =  	[[3A]] 	 +  	[[0A]] 	* 	x_1
>>>

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

   <<<
 POL(a(x_1, x_2)) =  	[[2A]] 	 +  	[[0A]] 	* 	x_1 	 +  	[[2A]] 	* 	x_2
>>>

   <<<
 POL(b(x_1, x_2)) =  	[[4A]] 	 +  	[[1A]] 	* 	x_1 	 +  	[[-I]] 	* 	x_2
>>>

   <<<
 POL(0) =  	[[0A]]
>>>


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

   c(c(c(a(x, y)))) -> b(c(c(c(c(y)))), x)
   c(c(b(c(y), 0))) -> a(0, c(c(a(y, 0))))
   c(c(a(a(y, 0), x))) -> c(y)


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

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

   c(c(c(a(x, y)))) -> b(c(c(c(c(y)))), x)
   c(c(b(c(y), 0))) -> a(0, c(c(a(y, 0))))
   c(c(a(a(y, 0), x))) -> c(y)

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