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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 34 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) MRRProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPOrderProof [EQUIVALENT, 15 ms]
        (9) QDP
        (10) PisEmptyProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) MRRProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) QDPOrderProof [EQUIVALENT, 461 ms]
        (16) QDP
        (17) PisEmptyProof [EQUIVALENT, 0 ms]
        (18) YES


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

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

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

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:

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

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

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

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

(6) 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:

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


Used ordering: Polynomial interpretation [POLO]:

   POL(F(x_1, x_2)) = 2*x_1 + x_2
   POL(a) = 1
   POL(b) = 0
   POL(f(x_1, x_2)) = x_1 + 2*x_2


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

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

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

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

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

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

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

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

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


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

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


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

(9)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

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

(11)
YES

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

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

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

The TRS R consists of the following rules:

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

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

(13) 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:

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


Used ordering: Polynomial interpretation [POLO]:

   POL(F(x_1, x_2)) = x_1 + x_2
   POL(a) = 2
   POL(b) = 0
   POL(f(x_1, x_2)) = x_1 + x_2


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

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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


The following pairs can be oriented strictly and are deleted.

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

   <<<
 POL(F(x_1, x_2)) =  	[[0A]] 	 +  	[[-I, -I, 0A]] 	* 	x_1 	 +  	[[-I, -I, 0A]] 	* 	x_2
>>>

   <<<
 POL(a) =  	[[0A], [0A], [-I]]
>>>

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

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


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

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


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

(16)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

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

(18)
YES