YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z119.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 9 ms)
↳2 QDP
↳3 MRRProof (⇔, 51 ms)
↳4 QDP
↳5 MRRProof (⇔, 0 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 QDP
↳9 QDPOrderProof (⇔, 14 ms)
↳10 QDP
↳11 PisEmptyProof (⇔, 0 ms)
↳12 YES

(0) Obligation:

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

a(b(x)) → b(d(x))
a(c(x)) → d(d(d(x)))
b(d(x)) → a(c(b(x)))
c(f(x)) → d(d(c(x)))
d(d(x)) → f(x)
f(f(x)) → a(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(b(x)) → B(d(x))
A(b(x)) → D(x)
A(c(x)) → D(d(d(x)))
A(c(x)) → D(d(x))
A(c(x)) → D(x)
B(d(x)) → A(c(b(x)))
B(d(x)) → C(b(x))
B(d(x)) → B(x)
C(f(x)) → D(d(c(x)))
C(f(x)) → D(c(x))
C(f(x)) → C(x)
D(d(x)) → F(x)
F(f(x)) → A(x)

The TRS R consists of the following rules:

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

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

(3) 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:

A(b(x)) → D(x)


Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = x1   
POL(B(x1)) = 1 + 2·x1   
POL(C(x1)) = x1   
POL(D(x1)) = x1   
POL(F(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = 1 + 2·x1   
POL(c(x1)) = x1   
POL(d(x1)) = x1   
POL(f(x1)) = x1   

(4) Obligation:

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

A(b(x)) → B(d(x))
A(c(x)) → D(d(d(x)))
A(c(x)) → D(d(x))
A(c(x)) → D(x)
B(d(x)) → A(c(b(x)))
B(d(x)) → C(b(x))
B(d(x)) → B(x)
C(f(x)) → D(d(c(x)))
C(f(x)) → D(c(x))
C(f(x)) → C(x)
D(d(x)) → F(x)
F(f(x)) → A(x)

The TRS R consists of the following rules:

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

A(c(x)) → D(d(d(x)))
A(c(x)) → D(d(x))
A(c(x)) → D(x)
B(d(x)) → C(b(x))
B(d(x)) → B(x)
C(f(x)) → D(d(c(x)))
C(f(x)) → D(c(x))
C(f(x)) → C(x)

Strictly oriented rules of the TRS R:

f(f(x)) → a(x)

Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = 3 + x1   
POL(B(x1)) = 3·x1   
POL(C(x1)) = 2 + x1   
POL(D(x1)) = x1   
POL(F(x1)) = 1 + x1   
POL(a(x1)) = 3 + x1   
POL(b(x1)) = 3·x1   
POL(c(x1)) = x1   
POL(d(x1)) = 1 + x1   
POL(f(x1)) = 2 + x1   

(6) Obligation:

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

A(b(x)) → B(d(x))
B(d(x)) → A(c(b(x)))
D(d(x)) → F(x)
F(f(x)) → A(x)

The TRS R consists of the following rules:

a(b(x)) → b(d(x))
a(c(x)) → d(d(d(x)))
b(d(x)) → a(c(b(x)))
c(f(x)) → d(d(c(x)))
d(d(x)) → f(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 1 SCC with 2 less nodes.

(8) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(9) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL(A(x1)) = 4·x1   
POL(B(x1)) = 5 + 5·x1   
POL(a(x1)) = 3   
POL(b(x1)) = 4 + x1   
POL(c(x1)) = 2   
POL(d(x1)) = 2   
POL(f(x1)) = 0   

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

c(f(x)) → d(d(c(x)))
d(d(x)) → f(x)

(10) Obligation:

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

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

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

(11) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(12) YES