YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z111.srs

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

0 QTRS
↳1 QTRSRRRProof (⇔, 57 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 18 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 QDP
↳7 QDPOrderProof (⇔, 14 ms)
↳8 QDP
↳9 PisEmptyProof (⇔, 0 ms)
↳10 YES

(0) Obligation:

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

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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a(x1)) = 25 + x1   
POL(b(x1)) = 17 + x1   
POL(c(x1)) = 11 + x1   
POL(d(x1)) = 7 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

b(b(x)) → c(c(c(x)))
c(c(x)) → d(d(d(x)))


(2) Obligation:

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

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

The TRS R consists of the following rules:

a(a(x)) → b(c(c(c(x))))
b(c(x)) → d(d(d(d(x))))
a(x) → d(c(d(x)))
c(d(d(x))) → a(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 1 less node.

(6) Obligation:

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

A(a(x)) → C(c(c(x)))
C(d(d(x))) → A(x)
A(a(x)) → C(c(x))
A(a(x)) → C(x)
A(x) → C(d(x))

The TRS R consists of the following rules:

a(a(x)) → b(c(c(c(x))))
b(c(x)) → d(d(d(d(x))))
a(x) → d(c(d(x)))
c(d(d(x))) → a(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.


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

POL(A(x1)) = 4 + 2·x1   
POL(C(x1)) = 1 + 2·x1   
POL(a(x1)) = 5 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 3 + x1   
POL(d(x1)) = 1 + x1   

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

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

(8) Obligation:

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

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