YES
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 20 ms]
(4) QTRS
(5) DependencyPairsProof [EQUIVALENT, 1 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) UsableRulesProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) DependencyGraphProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) MNOCProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) UsableRulesProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) QReductionProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (21) YES
    (22) QDP
        (23) UsableRulesProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) MNOCProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) MRRProof [EQUIVALENT, 0 ms]
        (28) QDP
        (29) PisEmptyProof [EQUIVALENT, 0 ms]
        (30) YES
    (31) QDP
        (32) UsableRulesProof [EQUIVALENT, 0 ms]
        (33) QDP
        (34) QDPOrderProof [EQUIVALENT, 22 ms]
        (35) QDP
        (36) DependencyGraphProof [EQUIVALENT, 0 ms]
        (37) QDP
        (38) QDPOrderProof [EQUIVALENT, 12 ms]
        (39) QDP
        (40) QDPOrderProof [EQUIVALENT, 340 ms]
        (41) QDP
        (42) QDPOrderProof [EQUIVALENT, 200 ms]
        (43) QDP
        (44) DependencyGraphProof [EQUIVALENT, 0 ms]
        (45) TRUE


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

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

   a(b(x1)) -> b(c(a(x1)))
   b(c(x1)) -> c(b(b(x1)))
   a(c(x1)) -> c(a(b(x1)))
   a(a(x1)) -> a(d(d(d(x1))))
   d(a(x1)) -> d(d(c(x1)))
   a(d(d(c(x1)))) -> a(a(a(d(x1))))
   e(e(f(f(x1)))) -> f(f(f(e(e(x1)))))
   e(x1) -> a(x1)
   b(d(x1)) -> d(d(x1))

Q is empty.

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

(1) QTRS Reverse (EQUIVALENT)
We applied the QTRS Reverse Processor [REVERSE].
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   b(a(x1)) -> a(c(b(x1)))
   c(b(x1)) -> b(b(c(x1)))
   c(a(x1)) -> b(a(c(x1)))
   a(a(x1)) -> d(d(d(a(x1))))
   a(d(x1)) -> c(d(d(x1)))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))
   f(f(e(e(x1)))) -> e(e(f(f(f(x1)))))
   e(x1) -> a(x1)
   d(b(x1)) -> d(d(x1))

Q is empty.

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

(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(a(x_1)) = x_1
   POL(b(x_1)) = x_1
   POL(c(x_1)) = x_1
   POL(d(x_1)) = x_1
   POL(e(x_1)) = 1 + x_1
   POL(f(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   e(x1) -> a(x1)




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

(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   b(a(x1)) -> a(c(b(x1)))
   c(b(x1)) -> b(b(c(x1)))
   c(a(x1)) -> b(a(c(x1)))
   a(a(x1)) -> d(d(d(a(x1))))
   a(d(x1)) -> c(d(d(x1)))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))
   f(f(e(e(x1)))) -> e(e(f(f(f(x1)))))
   d(b(x1)) -> d(d(x1))

Q is empty.

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

(5) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

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

   B(a(x1)) -> A(c(b(x1)))
   B(a(x1)) -> C(b(x1))
   B(a(x1)) -> B(x1)
   C(b(x1)) -> B(b(c(x1)))
   C(b(x1)) -> B(c(x1))
   C(b(x1)) -> C(x1)
   C(a(x1)) -> B(a(c(x1)))
   C(a(x1)) -> A(c(x1))
   C(a(x1)) -> C(x1)
   A(a(x1)) -> D(d(d(a(x1))))
   A(a(x1)) -> D(d(a(x1)))
   A(a(x1)) -> D(a(x1))
   A(d(x1)) -> C(d(d(x1)))
   A(d(x1)) -> D(d(x1))
   C(d(d(a(x1)))) -> D(a(a(a(x1))))
   C(d(d(a(x1)))) -> A(a(a(x1)))
   C(d(d(a(x1)))) -> A(a(x1))
   F(f(e(e(x1)))) -> F(f(f(x1)))
   F(f(e(e(x1)))) -> F(f(x1))
   F(f(e(e(x1)))) -> F(x1)
   D(b(x1)) -> D(d(x1))
   D(b(x1)) -> D(x1)

The TRS R consists of the following rules:

   b(a(x1)) -> a(c(b(x1)))
   c(b(x1)) -> b(b(c(x1)))
   c(a(x1)) -> b(a(c(x1)))
   a(a(x1)) -> d(d(d(a(x1))))
   a(d(x1)) -> c(d(d(x1)))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))
   f(f(e(e(x1)))) -> e(e(f(f(f(x1)))))
   d(b(x1)) -> d(d(x1))

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

(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 5 less nodes.
----------------------------------------

(8)
Complex Obligation (AND)

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

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

   D(b(x1)) -> D(x1)
   D(b(x1)) -> D(d(x1))

The TRS R consists of the following rules:

   b(a(x1)) -> a(c(b(x1)))
   c(b(x1)) -> b(b(c(x1)))
   c(a(x1)) -> b(a(c(x1)))
   a(a(x1)) -> d(d(d(a(x1))))
   a(d(x1)) -> c(d(d(x1)))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))
   f(f(e(e(x1)))) -> e(e(f(f(f(x1)))))
   d(b(x1)) -> d(d(x1))

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

(10) 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.
----------------------------------------

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   D(b(x1)) -> D(x1)
   D(b(x1)) -> D(d(x1))

The TRS R consists of the following rules:

   d(b(x1)) -> d(d(x1))

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

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

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   D(b(x1)) -> D(x1)

The TRS R consists of the following rules:

   d(b(x1)) -> d(d(x1))

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

(14) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   D(b(x1)) -> D(x1)

The TRS R consists of the following rules:

   d(b(x1)) -> d(d(x1))

The set Q consists of the following terms:

   d(b(x0))

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

(16) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(17)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   D(b(x1)) -> D(x1)

R is empty.
The set Q consists of the following terms:

   d(b(x0))

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

(18) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   d(b(x0))


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

(19)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   D(b(x1)) -> D(x1)

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

(20) 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:
*D(b(x1)) -> D(x1)
The graph contains the following edges 1 > 1


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

(21)
YES

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

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

   F(f(e(e(x1)))) -> F(f(x1))
   F(f(e(e(x1)))) -> F(f(f(x1)))
   F(f(e(e(x1)))) -> F(x1)

The TRS R consists of the following rules:

   b(a(x1)) -> a(c(b(x1)))
   c(b(x1)) -> b(b(c(x1)))
   c(a(x1)) -> b(a(c(x1)))
   a(a(x1)) -> d(d(d(a(x1))))
   a(d(x1)) -> c(d(d(x1)))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))
   f(f(e(e(x1)))) -> e(e(f(f(f(x1)))))
   d(b(x1)) -> d(d(x1))

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

(23) 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.
----------------------------------------

(24)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(f(e(e(x1)))) -> F(f(x1))
   F(f(e(e(x1)))) -> F(f(f(x1)))
   F(f(e(e(x1)))) -> F(x1)

The TRS R consists of the following rules:

   f(f(e(e(x1)))) -> e(e(f(f(f(x1)))))

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

(25) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

(26)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(f(e(e(x1)))) -> F(f(x1))
   F(f(e(e(x1)))) -> F(f(f(x1)))
   F(f(e(e(x1)))) -> F(x1)

The TRS R consists of the following rules:

   f(f(e(e(x1)))) -> e(e(f(f(f(x1)))))

The set Q consists of the following terms:

   f(f(e(e(x0))))

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

(27) 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(e(e(x1)))) -> F(f(x1))
   F(f(e(e(x1)))) -> F(f(f(x1)))
   F(f(e(e(x1)))) -> F(x1)


Used ordering: Polynomial interpretation [POLO]:

   POL(F(x_1)) = x_1
   POL(e(x_1)) = 2 + x_1
   POL(f(x_1)) = x_1


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

(28)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   f(f(e(e(x1)))) -> e(e(f(f(f(x1)))))

The set Q consists of the following terms:

   f(f(e(e(x0))))

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

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

(30)
YES

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

(31)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(d(x1)) -> C(d(d(x1)))
   C(b(x1)) -> B(b(c(x1)))
   B(a(x1)) -> A(c(b(x1)))
   B(a(x1)) -> C(b(x1))
   C(b(x1)) -> B(c(x1))
   B(a(x1)) -> B(x1)
   C(b(x1)) -> C(x1)
   C(a(x1)) -> B(a(c(x1)))
   C(a(x1)) -> A(c(x1))
   C(a(x1)) -> C(x1)
   C(d(d(a(x1)))) -> A(a(a(x1)))
   C(d(d(a(x1)))) -> A(a(x1))

The TRS R consists of the following rules:

   b(a(x1)) -> a(c(b(x1)))
   c(b(x1)) -> b(b(c(x1)))
   c(a(x1)) -> b(a(c(x1)))
   a(a(x1)) -> d(d(d(a(x1))))
   a(d(x1)) -> c(d(d(x1)))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))
   f(f(e(e(x1)))) -> e(e(f(f(f(x1)))))
   d(b(x1)) -> d(d(x1))

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

(32) 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.
----------------------------------------

(33)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(d(x1)) -> C(d(d(x1)))
   C(b(x1)) -> B(b(c(x1)))
   B(a(x1)) -> A(c(b(x1)))
   B(a(x1)) -> C(b(x1))
   C(b(x1)) -> B(c(x1))
   B(a(x1)) -> B(x1)
   C(b(x1)) -> C(x1)
   C(a(x1)) -> B(a(c(x1)))
   C(a(x1)) -> A(c(x1))
   C(a(x1)) -> C(x1)
   C(d(d(a(x1)))) -> A(a(a(x1)))
   C(d(d(a(x1)))) -> A(a(x1))

The TRS R consists of the following rules:

   a(a(x1)) -> d(d(d(a(x1))))
   a(d(x1)) -> c(d(d(x1)))
   c(b(x1)) -> b(b(c(x1)))
   b(a(x1)) -> a(c(b(x1)))
   c(a(x1)) -> b(a(c(x1)))
   d(b(x1)) -> d(d(x1))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))

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

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


The following pairs can be oriented strictly and are deleted.

   B(a(x1)) -> A(c(b(x1)))
   B(a(x1)) -> C(b(x1))
   B(a(x1)) -> B(x1)
   C(a(x1)) -> A(c(x1))
   C(a(x1)) -> C(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(A(x_1)) = 0
   POL(B(x_1)) = x_1
   POL(C(x_1)) = x_1
   POL(a(x_1)) = 1 + x_1
   POL(b(x_1)) = x_1
   POL(c(x_1)) = x_1
   POL(d(x_1)) = 0

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

   d(b(x1)) -> d(d(x1))
   a(d(x1)) -> c(d(d(x1)))
   c(b(x1)) -> b(b(c(x1)))
   b(a(x1)) -> a(c(b(x1)))
   c(a(x1)) -> b(a(c(x1)))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))
   a(a(x1)) -> d(d(d(a(x1))))


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

(35)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(d(x1)) -> C(d(d(x1)))
   C(b(x1)) -> B(b(c(x1)))
   C(b(x1)) -> B(c(x1))
   C(b(x1)) -> C(x1)
   C(a(x1)) -> B(a(c(x1)))
   C(d(d(a(x1)))) -> A(a(a(x1)))
   C(d(d(a(x1)))) -> A(a(x1))

The TRS R consists of the following rules:

   a(a(x1)) -> d(d(d(a(x1))))
   a(d(x1)) -> c(d(d(x1)))
   c(b(x1)) -> b(b(c(x1)))
   b(a(x1)) -> a(c(b(x1)))
   c(a(x1)) -> b(a(c(x1)))
   d(b(x1)) -> d(d(x1))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))

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

(36) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
----------------------------------------

(37)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C(b(x1)) -> C(x1)
   C(d(d(a(x1)))) -> A(a(a(x1)))
   A(d(x1)) -> C(d(d(x1)))
   C(d(d(a(x1)))) -> A(a(x1))

The TRS R consists of the following rules:

   a(a(x1)) -> d(d(d(a(x1))))
   a(d(x1)) -> c(d(d(x1)))
   c(b(x1)) -> b(b(c(x1)))
   b(a(x1)) -> a(c(b(x1)))
   c(a(x1)) -> b(a(c(x1)))
   d(b(x1)) -> d(d(x1))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))

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

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


The following pairs can be oriented strictly and are deleted.

   C(b(x1)) -> C(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(A(x_1)) = 0
   POL(C(x_1)) = x_1
   POL(a(x_1)) = 0
   POL(b(x_1)) = 1 + x_1
   POL(c(x_1)) = 0
   POL(d(x_1)) = 0

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

   d(b(x1)) -> d(d(x1))


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

(39)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C(d(d(a(x1)))) -> A(a(a(x1)))
   A(d(x1)) -> C(d(d(x1)))
   C(d(d(a(x1)))) -> A(a(x1))

The TRS R consists of the following rules:

   a(a(x1)) -> d(d(d(a(x1))))
   a(d(x1)) -> c(d(d(x1)))
   c(b(x1)) -> b(b(c(x1)))
   b(a(x1)) -> a(c(b(x1)))
   c(a(x1)) -> b(a(c(x1)))
   d(b(x1)) -> d(d(x1))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))

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

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


The following pairs can be oriented strictly and are deleted.

   C(d(d(a(x1)))) -> A(a(a(x1)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

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

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

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

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

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

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


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

   a(a(x1)) -> d(d(d(a(x1))))
   a(d(x1)) -> c(d(d(x1)))
   c(b(x1)) -> b(b(c(x1)))
   b(a(x1)) -> a(c(b(x1)))
   c(a(x1)) -> b(a(c(x1)))
   d(b(x1)) -> d(d(x1))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))


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

(41)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(d(x1)) -> C(d(d(x1)))
   C(d(d(a(x1)))) -> A(a(x1))

The TRS R consists of the following rules:

   a(a(x1)) -> d(d(d(a(x1))))
   a(d(x1)) -> c(d(d(x1)))
   c(b(x1)) -> b(b(c(x1)))
   b(a(x1)) -> a(c(b(x1)))
   c(a(x1)) -> b(a(c(x1)))
   d(b(x1)) -> d(d(x1))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))

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

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


The following pairs can be oriented strictly and are deleted.

   A(d(x1)) -> C(d(d(x1)))
The remaining pairs can at least be oriented weakly.
Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

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

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

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

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

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

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


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

   d(b(x1)) -> d(d(x1))
   a(a(x1)) -> d(d(d(a(x1))))
   a(d(x1)) -> c(d(d(x1)))
   c(b(x1)) -> b(b(c(x1)))
   b(a(x1)) -> a(c(b(x1)))
   c(a(x1)) -> b(a(c(x1)))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))


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

(43)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C(d(d(a(x1)))) -> A(a(x1))

The TRS R consists of the following rules:

   a(a(x1)) -> d(d(d(a(x1))))
   a(d(x1)) -> c(d(d(x1)))
   c(b(x1)) -> b(b(c(x1)))
   b(a(x1)) -> a(c(b(x1)))
   c(a(x1)) -> b(a(c(x1)))
   d(b(x1)) -> d(d(x1))
   c(d(d(a(x1)))) -> d(a(a(a(x1))))

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

(44) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(45)
TRUE