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


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YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 15 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 25 ms]
(6) QDP
(7) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(8) YES


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

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

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   B(f(b(x, z)), y) -> B(y, z)
   B(f(b(x, z)), y) -> B(z, b(y, z))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

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

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

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

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

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

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


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

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


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

   B(f(b(x, z)), y) -> B(z, b(y, z))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*B(f(b(x, z)), y) -> B(z, b(y, z))
The graph contains the following edges 1 > 1


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(8)
YES