YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z069.srs

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

0 QTRS
↳1 QTRSRRRProof (⇔, 48 ms)
↳2 QTRS
↳3 Overlay + Local Confluence (⇔, 0 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 15 ms)
↳6 QDP
↳7 QDPOrderProof (⇔, 44 ms)
↳8 QDP
↳9 PisEmptyProof (⇔, 0 ms)
↳10 YES

(0) Obligation:

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

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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(A(x1)) = 1 + x1   
POL(B(x1)) = 1 + x1   
POL(C(x1)) = 1 + x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 1 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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


(2) Obligation:

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

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

Q is empty.

(3) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(4) Obligation:

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

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

The set Q consists of the following terms:

B(a(a(a(x0))))
A(A(A(b(x0))))
C(b(b(b(x0))))
B(B(B(c(x0))))
A(c(c(c(x0))))
C(C(C(a(x0))))

(5) DependencyPairsProof (EQUIVALENT transformation)

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:

B1(a(a(a(x)))) → A1(A(A(x)))
B1(a(a(a(x)))) → A1(A(x))
B1(a(a(a(x)))) → A1(x)
A1(A(A(b(x)))) → C1(x)
C1(b(b(b(x)))) → B1(B(B(x)))
C1(b(b(b(x)))) → B1(B(x))
C1(b(b(b(x)))) → B1(x)
B1(B(B(c(x)))) → A1(x)
A1(c(c(c(x)))) → C1(C(C(x)))
A1(c(c(c(x)))) → C1(C(x))
A1(c(c(c(x)))) → C1(x)
C1(C(C(a(x)))) → B1(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

B(a(a(a(x0))))
A(A(A(b(x0))))
C(b(b(b(x0))))
B(B(B(c(x0))))
A(c(c(c(x0))))
C(C(C(a(x0))))

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.


B1(a(a(a(x)))) → A1(A(A(x)))
B1(a(a(a(x)))) → A1(A(x))
B1(a(a(a(x)))) → A1(x)
A1(A(A(b(x)))) → C1(x)
C1(b(b(b(x)))) → B1(B(B(x)))
C1(b(b(b(x)))) → B1(B(x))
C1(b(b(b(x)))) → B1(x)
B1(B(B(c(x)))) → A1(x)
A1(c(c(c(x)))) → C1(C(C(x)))
A1(c(c(c(x)))) → C1(C(x))
A1(c(c(c(x)))) → C1(x)
C1(C(C(a(x)))) → B1(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = 1 + x1   
POL(A1(x1)) = 1 + x1   
POL(B(x1)) = 1 + x1   
POL(B1(x1)) = 1 + x1   
POL(C(x1)) = 1 + x1   
POL(C1(x1)) = 1 + 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:

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

(8) Obligation:

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

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

The set Q consists of the following terms:

B(a(a(a(x0))))
A(A(A(b(x0))))
C(b(b(b(x0))))
B(B(B(c(x0))))
A(c(c(c(x0))))
C(C(C(a(x0))))

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