YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_06/beans4.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 14 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 25 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 TRUE

(0) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(6) Obligation:

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

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

The TRS R consists of the following rules:

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