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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 21 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 48 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 46 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) QDP
(11) UsableRulesProof [EQUIVALENT, 0 ms]
(12) QDP
(13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(14) YES


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

   a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
   f(a(x), a(y)) -> a(f(x, y))
   f(b(x), b(y)) -> b(f(x, 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:

   A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
   A(a(f(x, y))) -> A(b(a(b(a(x)))))
   A(a(f(x, y))) -> A(b(a(x)))
   A(a(f(x, y))) -> A(x)
   A(a(f(x, y))) -> A(b(a(b(a(y)))))
   A(a(f(x, y))) -> A(b(a(y)))
   A(a(f(x, y))) -> A(y)
   F(a(x), a(y)) -> A(f(x, y))
   F(a(x), a(y)) -> F(x, y)
   F(b(x), b(y)) -> F(x, y)

The TRS R consists of the following rules:

   a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
   f(a(x), a(y)) -> a(f(x, y))
   f(b(x), b(y)) -> b(f(x, 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 4 less nodes.
----------------------------------------

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

   F(a(x), a(y)) -> A(f(x, y))
   A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
   F(a(x), a(y)) -> F(x, y)
   F(b(x), b(y)) -> F(x, y)
   A(a(f(x, y))) -> A(x)
   A(a(f(x, y))) -> A(y)

The TRS R consists of the following rules:

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

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

(5) 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:

   A(a(f(x, y))) -> A(x)
   A(a(f(x, y))) -> A(y)


Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = 2*x_1
   POL(F(x_1, x_2)) = 2 + 2*x_1 + 2*x_2
   POL(a(x_1)) = x_1
   POL(b(x_1)) = x_1
   POL(f(x_1, x_2)) = 1 + x_1 + x_2


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

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

   F(a(x), a(y)) -> A(f(x, y))
   A(a(f(x, y))) -> F(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
   F(a(x), a(y)) -> F(x, y)
   F(b(x), b(y)) -> F(x, y)

The TRS R consists of the following rules:

   a(a(f(x, y))) -> f(a(b(a(b(a(x))))), a(b(a(b(a(y))))))
   f(a(x), a(y)) -> a(f(x, y))
   f(b(x), b(y)) -> b(f(x, 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.

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

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

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

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

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

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


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

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


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

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

   F(a(x), a(y)) -> A(f(x, y))
   F(a(x), a(y)) -> F(x, y)
   F(b(x), b(y)) -> F(x, y)

The TRS R consists of the following rules:

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

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

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

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

   F(b(x), b(y)) -> F(x, y)
   F(a(x), a(y)) -> F(x, y)

The TRS R consists of the following rules:

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

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

(11) 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.
----------------------------------------

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

   F(b(x), b(y)) -> F(x, y)
   F(a(x), a(y)) -> F(x, y)

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

(13) 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:
*F(b(x), b(y)) -> F(x, y)
The graph contains the following edges 1 > 1, 2 > 2


*F(a(x), a(y)) -> F(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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(14)
YES