YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z066.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 32 ms)
↳4 QTRS
↳5 QTRSRRRProof (⇔, 0 ms)
↳6 QTRS
↳7 QTRSRRRProof (⇔, 1 ms)
↳8 QTRS
↳9 DependencyPairsProof (⇔, 0 ms)
↳10 QDP
↳11 QDPOrderProof (⇔, 57 ms)
↳12 QDP
↳13 DependencyGraphProof (⇔, 0 ms)
↳14 AND
    ↳15 QDP
        ↳16 QDPOrderProof (⇔, 10 ms)
        ↳17 QDP
        ↳18 DependencyGraphProof (⇔, 0 ms)
        ↳19 TRUE
    ↳20 QDP
        ↳21 QDPOrderProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 DependencyGraphProof (⇔, 0 ms)
        ↳24 TRUE

(0) Obligation:

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

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

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

(3) QTRSRRRProof (EQUIVALENT transformation)

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

A(a(x)) → x
a(A(x)) → x


(4) Obligation:

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

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

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

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

B(b(x)) → x
b(B(x)) → x


(6) Obligation:

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

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

Q is empty.

(7) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(A(x1)) = x1   
POL(B(x1)) = x1   
POL(C(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = 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:

C(c(x)) → x
c(C(x)) → x


(8) Obligation:

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

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

Q is empty.

(9) DependencyPairsProof (EQUIVALENT transformation)

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

(10) Obligation:

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

C1(b(a(x))) → A1(b(c(x)))
C1(b(a(x))) → B1(c(x))
C1(b(a(x))) → C1(x)
A2(B(C(x))) → C2(B(A(x)))
A2(B(C(x))) → B2(A(x))
A2(B(C(x))) → A2(x)
C2(a(b(x))) → B1(a(C(x)))
C2(a(b(x))) → A1(C(x))
C2(a(b(x))) → C2(x)
B2(A(c(x))) → C1(A(B(x)))
B2(A(c(x))) → A2(B(x))
B2(A(c(x))) → B2(x)
B1(c(A(x))) → A2(c(b(x)))
B1(c(A(x))) → C1(b(x))
B1(c(A(x))) → B1(x)
A1(C(B(x))) → B2(C(a(x)))
A1(C(B(x))) → C2(a(x))
A1(C(B(x))) → A1(x)

The TRS R consists of the following rules:

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


C1(b(a(x))) → B1(c(x))
C1(b(a(x))) → C1(x)
A2(B(C(x))) → A2(x)
C2(a(b(x))) → C2(x)
B2(A(c(x))) → A2(B(x))
B2(A(c(x))) → B2(x)
B1(c(A(x))) → B1(x)
A1(C(B(x))) → C2(a(x))
A1(C(B(x))) → A1(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = 1 + x1   
POL(A1(x1)) = x1   
POL(A2(x1)) = x1   
POL(B(x1)) = x1   
POL(B1(x1)) = x1   
POL(B2(x1)) = x1   
POL(C(x1)) = 1 + x1   
POL(C1(x1)) = x1   
POL(C2(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = x1   

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

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

(12) Obligation:

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

C1(b(a(x))) → A1(b(c(x)))
A2(B(C(x))) → C2(B(A(x)))
A2(B(C(x))) → B2(A(x))
C2(a(b(x))) → B1(a(C(x)))
C2(a(b(x))) → A1(C(x))
B2(A(c(x))) → C1(A(B(x)))
B1(c(A(x))) → A2(c(b(x)))
B1(c(A(x))) → C1(b(x))
A1(C(B(x))) → B2(C(a(x)))

The TRS R consists of the following rules:

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

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

(13) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(14) Complex Obligation (AND)

(15) Obligation:

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

A1(C(B(x))) → B2(C(a(x)))
B2(A(c(x))) → C1(A(B(x)))
C1(b(a(x))) → A1(b(c(x)))

The TRS R consists of the following rules:

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

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

(16) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


B2(A(c(x))) → C1(A(B(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = 0   
POL(A1(x1)) = 1   
POL(B(x1)) = 0   
POL(B2(x1)) = 1   
POL(C(x1)) = x1   
POL(C1(x1)) = x1   
POL(a(x1)) = 1   
POL(b(x1)) = 1   
POL(c(x1)) = x1   

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

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

(17) Obligation:

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

A1(C(B(x))) → B2(C(a(x)))
C1(b(a(x))) → A1(b(c(x)))

The TRS R consists of the following rules:

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

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

(18) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(19) TRUE

(20) Obligation:

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

A2(B(C(x))) → C2(B(A(x)))
C2(a(b(x))) → B1(a(C(x)))
B1(c(A(x))) → A2(c(b(x)))

The TRS R consists of the following rules:

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

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

(21) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


C2(a(b(x))) → B1(a(C(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = 1   
POL(A2(x1)) = 1   
POL(B(x1)) = x1   
POL(B1(x1)) = x1   
POL(C(x1)) = 0   
POL(C2(x1)) = 1   
POL(a(x1)) = 0   
POL(b(x1)) = x1   
POL(c(x1)) = 1   

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

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

(22) Obligation:

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

A2(B(C(x))) → C2(B(A(x)))
B1(c(A(x))) → A2(c(b(x)))

The TRS R consists of the following rules:

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

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

(23) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(24) TRUE