YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-502.srs

(0) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(5) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL(A(x1)) = x1   
POL(B(x1)) = x1   
POL(C(x1)) = x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 1 + x1   

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

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

(6) Obligation:

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

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

The TRS R consists of the following rules:

a(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(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.


C(b(x)) → 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 +
[-I,-I,0A]
·x1

POL(a(x1)) =
/1A\
|-I|
\-I/
+
/-I0A1A\
|0A0A0A|
\-I0A0A/
·x1

POL(B(x1)) = 0A +
[-I,0A,0A]
·x1

POL(b(x1)) =
/1A\
|-I|
\-I/
+
/0A1A1A\
|-I0A0A|
\-I0A0A/
·x1

POL(C(x1)) = 0A +
[0A,0A,0A]
·x1

POL(c(x1)) =
/-I\
|-I|
\-I/
+
/0A0A0A\
|0A0A0A|
\0A0A0A/
·x1

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

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

(8) Obligation:

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

A(a(x)) → B(b(x))
B(a(x)) → C(c(x))

The TRS R consists of the following rules:

a(a(x)) → b(b(x))
b(a(x)) → c(c(x))
c(b(x)) → a(a(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 0 SCCs with 2 less nodes.

(10) TRUE