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-552.srs

(0) Obligation:

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

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

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

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

The TRS R consists of the following rules:

b(a(x)) → x
c(a(x)) → b(a(a(c(c(b(x))))))
c(b(x)) → 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 2 less nodes.

(6) Obligation:

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

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

The TRS R consists of the following rules:

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

POL(C(x1)) = x1   
POL(a(x1)) = [2] + [4]x1   
POL(b(x1)) = [1/4]x1   
POL(c(x1)) = [4]x1   
The value of delta used in the strict ordering is 2.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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

(8) Obligation:

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

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

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

(9) PisEmptyProof (EQUIVALENT transformation)

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

(10) YES