NO
proof of Transformed_CSR_04_Ex9_Luc04_iGM.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 2 ms]
        (9) YES
    (10) QDP
        (11) QDPOrderProof [EQUIVALENT, 39 ms]
        (12) QDP
        (13) QDPOrderProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) TransformationProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPOrderProof [EQUIVALENT, 1234 ms]
        (18) QDP
        (19) QDPOrderProof [EQUIVALENT, 868 ms]
        (20) QDP
        (21) QDPOrderProof [EQUIVALENT, 1799 ms]
        (22) QDP
        (23) NonTerminationLoopProof [COMPLETE, 9500 ms]
        (24) NO


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

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

   active(f(a, b, X)) -> mark(f(X, X, X))
   active(c) -> mark(a)
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

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:

   ACTIVE(f(a, b, X)) -> MARK(f(X, X, X))
   ACTIVE(f(a, b, X)) -> F(X, X, X)
   ACTIVE(c) -> MARK(a)
   ACTIVE(c) -> MARK(b)
   MARK(f(X1, X2, X3)) -> ACTIVE(f(mark(X1), X2, mark(X3)))
   MARK(f(X1, X2, X3)) -> F(mark(X1), X2, mark(X3))
   MARK(f(X1, X2, X3)) -> MARK(X1)
   MARK(f(X1, X2, X3)) -> MARK(X3)
   MARK(a) -> ACTIVE(a)
   MARK(b) -> ACTIVE(b)
   MARK(c) -> ACTIVE(c)
   F(mark(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, mark(X2), X3) -> F(X1, X2, X3)
   F(X1, X2, mark(X3)) -> F(X1, X2, X3)
   F(active(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, active(X2), X3) -> F(X1, X2, X3)
   F(X1, X2, active(X3)) -> F(X1, X2, X3)

The TRS R consists of the following rules:

   active(f(a, b, X)) -> mark(f(X, X, X))
   active(c) -> mark(a)
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

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 7 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

   F(X1, mark(X2), X3) -> F(X1, X2, X3)
   F(mark(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, X2, mark(X3)) -> F(X1, X2, X3)
   F(active(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, active(X2), X3) -> F(X1, X2, X3)
   F(X1, X2, active(X3)) -> F(X1, X2, X3)

The TRS R consists of the following rules:

   active(f(a, b, X)) -> mark(f(X, X, X))
   active(c) -> mark(a)
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

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

(6) 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.
----------------------------------------

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

   F(X1, mark(X2), X3) -> F(X1, X2, X3)
   F(mark(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, X2, mark(X3)) -> F(X1, X2, X3)
   F(active(X1), X2, X3) -> F(X1, X2, X3)
   F(X1, active(X2), X3) -> F(X1, X2, X3)
   F(X1, X2, active(X3)) -> F(X1, X2, X3)

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

(8) 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(X1, mark(X2), X3) -> F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3


*F(mark(X1), X2, X3) -> F(X1, X2, X3)
The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3


*F(X1, X2, mark(X3)) -> F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3


*F(active(X1), X2, X3) -> F(X1, X2, X3)
The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3


*F(X1, active(X2), X3) -> F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3


*F(X1, X2, active(X3)) -> F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3


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

(9)
YES

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

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

   MARK(f(X1, X2, X3)) -> ACTIVE(f(mark(X1), X2, mark(X3)))
   ACTIVE(f(a, b, X)) -> MARK(f(X, X, X))
   MARK(f(X1, X2, X3)) -> MARK(X1)
   MARK(f(X1, X2, X3)) -> MARK(X3)

The TRS R consists of the following rules:

   active(f(a, b, X)) -> mark(f(X, X, X))
   active(c) -> mark(a)
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

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.

   MARK(f(X1, X2, X3)) -> MARK(X3)
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(MARK(x_1)) =  	[[2A]] 	 +  	[[0A]] 	* 	x_1
>>>

   <<<
 POL(f(x_1, x_2, x_3)) =  	[[3A]] 	 +  	[[0A]] 	* 	x_1 	 +  	[[0A]] 	* 	x_2 	 +  	[[3A]] 	* 	x_3
>>>

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

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

   <<<
 POL(a) =  	[[3A]]
>>>

   <<<
 POL(b) =  	[[1A]]
>>>

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

   <<<
 POL(c) =  	[[3A]]
>>>


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

   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   active(f(a, b, X)) -> mark(f(X, X, X))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)
   active(c) -> mark(a)
   active(c) -> mark(b)


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

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

   MARK(f(X1, X2, X3)) -> ACTIVE(f(mark(X1), X2, mark(X3)))
   ACTIVE(f(a, b, X)) -> MARK(f(X, X, X))
   MARK(f(X1, X2, X3)) -> MARK(X1)

The TRS R consists of the following rules:

   active(f(a, b, X)) -> mark(f(X, X, X))
   active(c) -> mark(a)
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

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

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


The following pairs can be oriented strictly and are deleted.

   MARK(f(X1, X2, X3)) -> MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(MARK(x_1)) =  	[[4A]] 	 +  	[[5A]] 	* 	x_1
>>>

   <<<
 POL(f(x_1, x_2, x_3)) =  	[[0A]] 	 +  	[[1A]] 	* 	x_1 	 +  	[[-I]] 	* 	x_2 	 +  	[[2A]] 	* 	x_3
>>>

   <<<
 POL(ACTIVE(x_1)) =  	[[4A]] 	 +  	[[5A]] 	* 	x_1
>>>

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

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

   <<<
 POL(b) =  	[[3A]]
>>>

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

   <<<
 POL(c) =  	[[3A]]
>>>


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

   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   active(f(a, b, X)) -> mark(f(X, X, X))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)
   active(c) -> mark(a)
   active(c) -> mark(b)


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

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

   MARK(f(X1, X2, X3)) -> ACTIVE(f(mark(X1), X2, mark(X3)))
   ACTIVE(f(a, b, X)) -> MARK(f(X, X, X))

The TRS R consists of the following rules:

   active(f(a, b, X)) -> mark(f(X, X, X))
   active(c) -> mark(a)
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

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

(15) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule MARK(f(X1, X2, X3)) -> ACTIVE(f(mark(X1), X2, mark(X3))) at position [0] we obtained the following new rules [LPAR04]:

   (MARK(f(x0, x1, y2)) -> ACTIVE(f(x0, x1, mark(y2))),MARK(f(x0, x1, y2)) -> ACTIVE(f(x0, x1, mark(y2))))
   (MARK(f(y0, mark(x1), y2)) -> ACTIVE(f(mark(y0), x1, mark(y2))),MARK(f(y0, mark(x1), y2)) -> ACTIVE(f(mark(y0), x1, mark(y2))))
   (MARK(f(y0, x1, x2)) -> ACTIVE(f(mark(y0), x1, x2)),MARK(f(y0, x1, x2)) -> ACTIVE(f(mark(y0), x1, x2)))
   (MARK(f(y0, active(x1), y2)) -> ACTIVE(f(mark(y0), x1, mark(y2))),MARK(f(y0, active(x1), y2)) -> ACTIVE(f(mark(y0), x1, mark(y2))))
   (MARK(f(f(x0, x1, x2), y1, y2)) -> ACTIVE(f(active(f(mark(x0), x1, mark(x2))), y1, mark(y2))),MARK(f(f(x0, x1, x2), y1, y2)) -> ACTIVE(f(active(f(mark(x0), x1, mark(x2))), y1, mark(y2))))
   (MARK(f(a, y1, y2)) -> ACTIVE(f(active(a), y1, mark(y2))),MARK(f(a, y1, y2)) -> ACTIVE(f(active(a), y1, mark(y2))))
   (MARK(f(b, y1, y2)) -> ACTIVE(f(active(b), y1, mark(y2))),MARK(f(b, y1, y2)) -> ACTIVE(f(active(b), y1, mark(y2))))
   (MARK(f(c, y1, y2)) -> ACTIVE(f(active(c), y1, mark(y2))),MARK(f(c, y1, y2)) -> ACTIVE(f(active(c), y1, mark(y2))))
   (MARK(f(y0, y1, f(x0, x1, x2))) -> ACTIVE(f(mark(y0), y1, active(f(mark(x0), x1, mark(x2))))),MARK(f(y0, y1, f(x0, x1, x2))) -> ACTIVE(f(mark(y0), y1, active(f(mark(x0), x1, mark(x2))))))
   (MARK(f(y0, y1, a)) -> ACTIVE(f(mark(y0), y1, active(a))),MARK(f(y0, y1, a)) -> ACTIVE(f(mark(y0), y1, active(a))))
   (MARK(f(y0, y1, b)) -> ACTIVE(f(mark(y0), y1, active(b))),MARK(f(y0, y1, b)) -> ACTIVE(f(mark(y0), y1, active(b))))
   (MARK(f(y0, y1, c)) -> ACTIVE(f(mark(y0), y1, active(c))),MARK(f(y0, y1, c)) -> ACTIVE(f(mark(y0), y1, active(c))))


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

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

   ACTIVE(f(a, b, X)) -> MARK(f(X, X, X))
   MARK(f(x0, x1, y2)) -> ACTIVE(f(x0, x1, mark(y2)))
   MARK(f(y0, mark(x1), y2)) -> ACTIVE(f(mark(y0), x1, mark(y2)))
   MARK(f(y0, x1, x2)) -> ACTIVE(f(mark(y0), x1, x2))
   MARK(f(y0, active(x1), y2)) -> ACTIVE(f(mark(y0), x1, mark(y2)))
   MARK(f(f(x0, x1, x2), y1, y2)) -> ACTIVE(f(active(f(mark(x0), x1, mark(x2))), y1, mark(y2)))
   MARK(f(a, y1, y2)) -> ACTIVE(f(active(a), y1, mark(y2)))
   MARK(f(b, y1, y2)) -> ACTIVE(f(active(b), y1, mark(y2)))
   MARK(f(c, y1, y2)) -> ACTIVE(f(active(c), y1, mark(y2)))
   MARK(f(y0, y1, f(x0, x1, x2))) -> ACTIVE(f(mark(y0), y1, active(f(mark(x0), x1, mark(x2)))))
   MARK(f(y0, y1, a)) -> ACTIVE(f(mark(y0), y1, active(a)))
   MARK(f(y0, y1, b)) -> ACTIVE(f(mark(y0), y1, active(b)))
   MARK(f(y0, y1, c)) -> ACTIVE(f(mark(y0), y1, active(c)))

The TRS R consists of the following rules:

   active(f(a, b, X)) -> mark(f(X, X, X))
   active(c) -> mark(a)
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

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

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


The following pairs can be oriented strictly and are deleted.

   MARK(f(y0, y1, b)) -> ACTIVE(f(mark(y0), y1, active(b)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(ACTIVE(x_1)) =  	[[1A]] 	 +  	[[-I, 1A, 2A]] 	* 	x_1
>>>

   <<<
 POL(f(x_1, x_2, x_3)) =  	[[2A], [1A], [0A]] 	 +  	[[-I, 0A, 3A], [-I, -I, -I], [-I, -I, 0A]] 	* 	x_1 	 +  	[[-I, -I, -I], [-I, -I, -I], [-I, -I, -I]] 	* 	x_2 	 +  	[[-I, -I, -I], [-I, 2A, 3A], [-I, -I, 3A]] 	* 	x_3
>>>

   <<<
 POL(a) =  	[[1A], [-I], [2A]]
>>>

   <<<
 POL(b) =  	[[1A], [-I], [-I]]
>>>

   <<<
 POL(MARK(x_1)) =  	[[3A]] 	 +  	[[0A, 1A, 2A]] 	* 	x_1
>>>

   <<<
 POL(mark(x_1)) =  	[[1A], [-I], [-I]] 	 +  	[[3A, 3A, 3A], [-I, 0A, 0A], [-I, -I, 0A]] 	* 	x_1
>>>

   <<<
 POL(active(x_1)) =  	[[0A], [-I], [-I]] 	 +  	[[3A, 3A, 1A], [-I, 0A, 0A], [-I, -I, 0A]] 	* 	x_1
>>>

   <<<
 POL(c) =  	[[0A], [3A], [2A]]
>>>


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

   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)
   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   active(f(a, b, X)) -> mark(f(X, X, X))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   active(c) -> mark(a)
   active(c) -> mark(b)


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

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(f(a, b, X)) -> MARK(f(X, X, X))
   MARK(f(x0, x1, y2)) -> ACTIVE(f(x0, x1, mark(y2)))
   MARK(f(y0, mark(x1), y2)) -> ACTIVE(f(mark(y0), x1, mark(y2)))
   MARK(f(y0, x1, x2)) -> ACTIVE(f(mark(y0), x1, x2))
   MARK(f(y0, active(x1), y2)) -> ACTIVE(f(mark(y0), x1, mark(y2)))
   MARK(f(f(x0, x1, x2), y1, y2)) -> ACTIVE(f(active(f(mark(x0), x1, mark(x2))), y1, mark(y2)))
   MARK(f(a, y1, y2)) -> ACTIVE(f(active(a), y1, mark(y2)))
   MARK(f(b, y1, y2)) -> ACTIVE(f(active(b), y1, mark(y2)))
   MARK(f(c, y1, y2)) -> ACTIVE(f(active(c), y1, mark(y2)))
   MARK(f(y0, y1, f(x0, x1, x2))) -> ACTIVE(f(mark(y0), y1, active(f(mark(x0), x1, mark(x2)))))
   MARK(f(y0, y1, a)) -> ACTIVE(f(mark(y0), y1, active(a)))
   MARK(f(y0, y1, c)) -> ACTIVE(f(mark(y0), y1, active(c)))

The TRS R consists of the following rules:

   active(f(a, b, X)) -> mark(f(X, X, X))
   active(c) -> mark(a)
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

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

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


The following pairs can be oriented strictly and are deleted.

   MARK(f(y0, y1, a)) -> ACTIVE(f(mark(y0), y1, active(a)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(ACTIVE(x_1)) =  	[[3A]] 	 +  	[[-I, -I, 1A]] 	* 	x_1
>>>

   <<<
 POL(f(x_1, x_2, x_3)) =  	[[3A], [3A], [-I]] 	 +  	[[-I, -I, -I], [-I, -I, -I], [-I, -I, -I]] 	* 	x_1 	 +  	[[-I, 0A, 1A], [-I, 0A, -I], [-I, 1A, 2A]] 	* 	x_2 	 +  	[[-I, 0A, 1A], [-I, 0A, 1A], [-I, 2A, 3A]] 	* 	x_3
>>>

   <<<
 POL(a) =  	[[2A], [-I], [-I]]
>>>

   <<<
 POL(b) =  	[[0A], [3A], [3A]]
>>>

   <<<
 POL(MARK(x_1)) =  	[[0A]] 	 +  	[[3A, -I, -I]] 	* 	x_1
>>>

   <<<
 POL(mark(x_1)) =  	[[3A], [0A], [-I]] 	 +  	[[-I, -I, 0A], [-I, 0A, 1A], [-I, -I, -I]] 	* 	x_1
>>>

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

   <<<
 POL(c) =  	[[3A], [3A], [3A]]
>>>


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

   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)
   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   active(f(a, b, X)) -> mark(f(X, X, X))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   active(c) -> mark(a)
   active(c) -> mark(b)


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

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(f(a, b, X)) -> MARK(f(X, X, X))
   MARK(f(x0, x1, y2)) -> ACTIVE(f(x0, x1, mark(y2)))
   MARK(f(y0, mark(x1), y2)) -> ACTIVE(f(mark(y0), x1, mark(y2)))
   MARK(f(y0, x1, x2)) -> ACTIVE(f(mark(y0), x1, x2))
   MARK(f(y0, active(x1), y2)) -> ACTIVE(f(mark(y0), x1, mark(y2)))
   MARK(f(f(x0, x1, x2), y1, y2)) -> ACTIVE(f(active(f(mark(x0), x1, mark(x2))), y1, mark(y2)))
   MARK(f(a, y1, y2)) -> ACTIVE(f(active(a), y1, mark(y2)))
   MARK(f(b, y1, y2)) -> ACTIVE(f(active(b), y1, mark(y2)))
   MARK(f(c, y1, y2)) -> ACTIVE(f(active(c), y1, mark(y2)))
   MARK(f(y0, y1, f(x0, x1, x2))) -> ACTIVE(f(mark(y0), y1, active(f(mark(x0), x1, mark(x2)))))
   MARK(f(y0, y1, c)) -> ACTIVE(f(mark(y0), y1, active(c)))

The TRS R consists of the following rules:

   active(f(a, b, X)) -> mark(f(X, X, X))
   active(c) -> mark(a)
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

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

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


The following pairs can be oriented strictly and are deleted.

   MARK(f(y0, y1, f(x0, x1, x2))) -> ACTIVE(f(mark(y0), y1, active(f(mark(x0), x1, mark(x2)))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

   <<<
 POL(ACTIVE(x_1)) =  	[[0A]] 	 +  	[[-I, 3A, 2A]] 	* 	x_1
>>>

   <<<
 POL(f(x_1, x_2, x_3)) =  	[[0A], [-I], [3A]] 	 +  	[[-I, -I, -I], [1A, -I, -I], [2A, -I, -I]] 	* 	x_1 	 +  	[[-I, -I, -I], [-I, -I, -I], [-I, -I, -I]] 	* 	x_2 	 +  	[[-I, -I, -I], [2A, -I, -I], [2A, -I, -I]] 	* 	x_3
>>>

   <<<
 POL(a) =  	[[2A], [-I], [-I]]
>>>

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

   <<<
 POL(MARK(x_1)) =  	[[0A]] 	 +  	[[-I, 3A, 3A]] 	* 	x_1
>>>

   <<<
 POL(mark(x_1)) =  	[[-I], [-I], [2A]] 	 +  	[[0A, -I, -I], [-I, 1A, -I], [-I, 0A, 0A]] 	* 	x_1
>>>

   <<<
 POL(active(x_1)) =  	[[-I], [-I], [2A]] 	 +  	[[0A, -I, -I], [-I, 1A, -I], [-I, -I, 0A]] 	* 	x_1
>>>

   <<<
 POL(c) =  	[[2A], [-I], [-I]]
>>>


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

   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)
   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   active(f(a, b, X)) -> mark(f(X, X, X))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   active(c) -> mark(a)
   active(c) -> mark(b)


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

(22)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ACTIVE(f(a, b, X)) -> MARK(f(X, X, X))
   MARK(f(x0, x1, y2)) -> ACTIVE(f(x0, x1, mark(y2)))
   MARK(f(y0, mark(x1), y2)) -> ACTIVE(f(mark(y0), x1, mark(y2)))
   MARK(f(y0, x1, x2)) -> ACTIVE(f(mark(y0), x1, x2))
   MARK(f(y0, active(x1), y2)) -> ACTIVE(f(mark(y0), x1, mark(y2)))
   MARK(f(f(x0, x1, x2), y1, y2)) -> ACTIVE(f(active(f(mark(x0), x1, mark(x2))), y1, mark(y2)))
   MARK(f(a, y1, y2)) -> ACTIVE(f(active(a), y1, mark(y2)))
   MARK(f(b, y1, y2)) -> ACTIVE(f(active(b), y1, mark(y2)))
   MARK(f(c, y1, y2)) -> ACTIVE(f(active(c), y1, mark(y2)))
   MARK(f(y0, y1, c)) -> ACTIVE(f(mark(y0), y1, active(c)))

The TRS R consists of the following rules:

   active(f(a, b, X)) -> mark(f(X, X, X))
   active(c) -> mark(a)
   active(c) -> mark(b)
   mark(f(X1, X2, X3)) -> active(f(mark(X1), X2, mark(X3)))
   mark(a) -> active(a)
   mark(b) -> active(b)
   mark(c) -> active(c)
   f(mark(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, mark(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, mark(X3)) -> f(X1, X2, X3)
   f(active(X1), X2, X3) -> f(X1, X2, X3)
   f(X1, active(X2), X3) -> f(X1, X2, X3)
   f(X1, X2, active(X3)) -> f(X1, X2, X3)

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

(23) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = MARK(f(active(c), active(c), X3')) evaluates to  t =MARK(f(X3', X3', X3'))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [ ]
* Semiunifier: [X3' / active(c)]

--------------------------------------------------------------------------------
Rewriting sequence

MARK(f(active(c), active(c), active(c))) -> MARK(f(active(c), mark(b), active(c)))
with rule active(c) -> mark(b) at position [0,1] and matcher [ ]

MARK(f(active(c), mark(b), active(c))) -> MARK(f(mark(a), mark(b), active(c)))
with rule active(c) -> mark(a) at position [0,0] and matcher [ ]

MARK(f(mark(a), mark(b), active(c))) -> MARK(f(mark(a), b, active(c)))
with rule f(X1, mark(X2), X3') -> f(X1, X2, X3') at position [0] and matcher [X1 / mark(a), X2 / b, X3' / active(c)]

MARK(f(mark(a), b, active(c))) -> MARK(f(a, b, active(c)))
with rule f(mark(X1), X2, X3) -> f(X1, X2, X3) at position [0] and matcher [X1 / a, X2 / b, X3 / active(c)]

MARK(f(a, b, active(c))) -> ACTIVE(f(mark(a), b, active(c)))
with rule MARK(f(y0, x1, x2)) -> ACTIVE(f(mark(y0), x1, x2)) at position [] and matcher [y0 / a, x1 / b, x2 / active(c)]

ACTIVE(f(mark(a), b, active(c))) -> ACTIVE(f(a, b, active(c)))
with rule f(mark(X1), X2, X3) -> f(X1, X2, X3) at position [0] and matcher [X1 / a, X2 / b, X3 / active(c)]

ACTIVE(f(a, b, active(c))) -> MARK(f(active(c), active(c), active(c)))
with rule ACTIVE(f(a, b, X)) -> MARK(f(X, X, X))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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

(24)
NO