YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z110.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 1 ms)
↳2 QDP
↳3 QDPOrderProof (⇔, 41 ms)
↳4 QDP
↳5 MRRProof (⇔, 13 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 TRUE

(0) Obligation:

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

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(3) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


B(b(c(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)) = 1 + x1   
POL(a(x1)) = x1   
POL(b(x1)) = 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)) → a(b(x))
a(a(x)) → b(x)
b(b(c(x))) → c(a(x))
c(a(x)) → b(a(c(x)))
b(b(x)) → a(a(a(x)))

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(5) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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

Strictly oriented rules of the TRS R:

a(a(x)) → b(x)
c(a(x)) → b(a(c(x)))

Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = x1   
POL(B(x1)) = 2 + x1   
POL(C(x1)) = 2 + 3·x1   
POL(a(x1)) = 2 + x1   
POL(b(x1)) = 3 + x1   
POL(c(x1)) = 3 + 3·x1   

(6) Obligation:

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

A(a(x)) → B(x)
B(b(c(x))) → C(a(x))

The TRS R consists of the following rules:

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

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) TRUE