YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Trafo_06/hom01.srs

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a(a(b(d(b(d(a(x))))))) → a(a(c(a(a(b(d(x)))))))
a(a(c(x))) → c(c(a(a(x))))
c(c(c(x))) → b(d(c(b(d(x)))))

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a(d(b(d(b(a(a(x))))))) → d(b(a(a(c(a(a(x)))))))
c(a(a(x))) → a(a(c(c(x))))
c(c(c(x))) → d(b(c(d(b(x)))))

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A(d(b(d(b(a(a(x))))))) → A(a(c(a(a(x)))))
A(d(b(d(b(a(a(x))))))) → A(c(a(a(x))))
A(d(b(d(b(a(a(x))))))) → C(a(a(x)))
C(a(a(x))) → A(a(c(c(x))))
C(a(a(x))) → A(c(c(x)))
C(a(a(x))) → C(c(x))
C(a(a(x))) → C(x)
C(c(c(x))) → C(d(b(x)))

The TRS R consists of the following rules:

a(d(b(d(b(a(a(x))))))) → d(b(a(a(c(a(a(x)))))))
c(a(a(x))) → a(a(c(c(x))))
c(c(c(x))) → d(b(c(d(b(x)))))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A(d(b(d(b(a(a(x))))))) → A(c(a(a(x))))
A(d(b(d(b(a(a(x))))))) → A(a(c(a(a(x)))))
A(d(b(d(b(a(a(x))))))) → C(a(a(x)))
C(a(a(x))) → A(a(c(c(x))))
C(a(a(x))) → A(c(c(x)))
C(a(a(x))) → C(c(x))
C(a(a(x))) → C(x)

The TRS R consists of the following rules:

a(d(b(d(b(a(a(x))))))) → d(b(a(a(c(a(a(x)))))))
c(a(a(x))) → a(a(c(c(x))))
c(c(c(x))) → d(b(c(d(b(x)))))

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


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

POL(A(x1)) = 0A +
[0A,-I,0A]
·x1

POL(d(x1)) =
/0A\
|0A|
\0A/
+
/0A1A-I\
|0A0A0A|
\0A-I-I/
·x1

POL(b(x1)) =
/0A\
|0A|
\-I/
+
/-I-I-I\
|-I0A-I|
\-I0A0A/
·x1

POL(a(x1)) =
/0A\
|0A|
\1A/
+
/0A-I-I\
|0A-I-I|
\0A0A0A/
·x1

POL(c(x1)) =
/0A\
|0A|
\1A/
+
/0A0A0A\
|-I-I0A|
\-I0A0A/
·x1

POL(C(x1)) = 1A +
[0A,0A,0A]
·x1

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

a(d(b(d(b(a(a(x))))))) → d(b(a(a(c(a(a(x)))))))
c(a(a(x))) → a(a(c(c(x))))
c(c(c(x))) → d(b(c(d(b(x)))))

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

C(a(a(x))) → A(a(c(c(x))))
C(a(a(x))) → A(c(c(x)))
C(a(a(x))) → C(c(x))
C(a(a(x))) → C(x)

The TRS R consists of the following rules:

a(d(b(d(b(a(a(x))))))) → d(b(a(a(c(a(a(x)))))))
c(a(a(x))) → a(a(c(c(x))))
c(c(c(x))) → d(b(c(d(b(x)))))

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

C(a(a(x))) → C(x)
C(a(a(x))) → C(c(x))

The TRS R consists of the following rules:

a(d(b(d(b(a(a(x))))))) → d(b(a(a(c(a(a(x)))))))
c(a(a(x))) → a(a(c(c(x))))
c(c(c(x))) → d(b(c(d(b(x)))))

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


C(a(a(x))) → C(x)
C(a(a(x))) → C(c(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(C(x1)) = x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = x1   
POL(d(x1)) = 0   

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

c(a(a(x))) → a(a(c(c(x))))
c(c(c(x))) → d(b(c(d(b(x)))))
a(d(b(d(b(a(a(x))))))) → d(b(a(a(c(a(a(x)))))))

(12) Obligation:

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

a(d(b(d(b(a(a(x))))))) → d(b(a(a(c(a(a(x)))))))
c(a(a(x))) → a(a(c(c(x))))
c(c(c(x))) → d(b(c(d(b(x)))))

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

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) YES