/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) QDPOrderProof [EQUIVALENT, 2 ms]
        (7) QDP
        (8) PisEmptyProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) QDPOrderProof [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:

   c(c(z, y, a), a, a) -> b(z, y)
   f(c(x, y, z)) -> c(z, f(b(y, z)), a)
   b(z, b(c(a, y, a), f(f(x)))) -> c(c(y, a, z), z, x)

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(z, y, a), a, a) -> B(z, y)
   F(c(x, y, z)) -> C(z, f(b(y, z)), a)
   F(c(x, y, z)) -> F(b(y, z))
   F(c(x, y, z)) -> B(y, z)
   B(z, b(c(a, y, a), f(f(x)))) -> C(c(y, a, z), z, x)
   B(z, b(c(a, y, a), f(f(x)))) -> C(y, a, z)

The TRS R consists of the following rules:

   c(c(z, y, a), a, a) -> b(z, y)
   f(c(x, y, z)) -> c(z, f(b(y, z)), a)
   b(z, b(c(a, y, a), f(f(x)))) -> c(c(y, a, z), z, x)

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 2 SCCs with 2 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B(z, b(c(a, y, a), f(f(x)))) -> C(c(y, a, z), z, x)
   C(c(z, y, a), a, a) -> B(z, y)
   B(z, b(c(a, y, a), f(f(x)))) -> C(y, a, z)

The TRS R consists of the following rules:

   c(c(z, y, a), a, a) -> b(z, y)
   f(c(x, y, z)) -> c(z, f(b(y, z)), a)
   b(z, b(c(a, y, a), f(f(x)))) -> c(c(y, a, z), z, x)

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

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


The following pairs can be oriented strictly and are deleted.

   B(z, b(c(a, y, a), f(f(x)))) -> C(c(y, a, z), z, x)
   C(c(z, y, a), a, a) -> B(z, y)
   B(z, b(c(a, y, a), f(f(x)))) -> C(y, a, z)
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
B(x1, x2)  =  x2

b(x1, x2)  =  b(x1, x2)

c(x1, x2, x3)  =  c(x1, x2)

a  =  a

f(x1)  =  f

C(x1, x2, x3)  =  x1


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   a=4
   b_2=11
   c_2=8
   f=7

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

   b(z, b(c(a, y, a), f(f(x)))) -> c(c(y, a, z), z, x)
   c(c(z, y, a), a, a) -> b(z, y)


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

(7)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   c(c(z, y, a), a, a) -> b(z, y)
   f(c(x, y, z)) -> c(z, f(b(y, z)), a)
   b(z, b(c(a, y, a), f(f(x)))) -> c(c(y, a, z), z, x)

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

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

(9)
YES

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

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

   F(c(x, y, z)) -> F(b(y, z))

The TRS R consists of the following rules:

   c(c(z, y, a), a, a) -> b(z, y)
   f(c(x, y, z)) -> c(z, f(b(y, z)), a)
   b(z, b(c(a, y, a), f(f(x)))) -> c(c(y, a, z), z, x)

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

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


The following pairs can be oriented strictly and are deleted.

   F(c(x, y, z)) -> F(b(y, z))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

   <<<
 POL(F(x_1)) =  	[[0]] 	 +  	[[0, 1]] 	* 	x_1
>>>

   <<<
 POL(c(x_1, x_2, x_3)) =  	[[0], [1]] 	 +  	[[1, 0], [1, 0]] 	* 	x_1 	 +  	[[1, 0], [1, 0]] 	* 	x_2 	 +  	[[0, 0], [1, 1]] 	* 	x_3
>>>

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

   <<<
 POL(a) =  	[[1], [1]]
>>>

   <<<
 POL(f(x_1)) =  	[[0], [0]] 	 +  	[[1, 1], [1, 0]] 	* 	x_1
>>>


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

   b(z, b(c(a, y, a), f(f(x)))) -> c(c(y, a, z), z, x)
   c(c(z, y, a), a, a) -> b(z, y)


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

(12)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   c(c(z, y, a), a, a) -> b(z, y)
   f(c(x, y, z)) -> c(z, f(b(y, z)), a)
   b(z, b(c(a, y, a), f(f(x)))) -> c(c(y, a, z), z, x)

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