YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_06/10.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 12 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 16 ms)
↳6 QDP
↳7 QDPOrderProof (⇔, 72 ms)
↳8 QDP
↳9 DependencyGraphProof (⇔, 0 ms)
↳10 TRUE

(0) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 6 less nodes.

(4) Obligation:

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

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

The TRS R consists of the following rules:

a(d(x)) → d(b(x))
a(x) → b(b(b(x)))
d(x) → x
a(x) → x
b(d(b(x))) → a(d(x))
b(c(x)) → c(d(d(x)))
a(c(x)) → b(b(c(d(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(d(x)) → B(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)) = x1   
POL(b(x1)) = x1   
POL(c(x1)) = 0   
POL(d(x1)) = 1 + x1   

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

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

(6) Obligation:

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

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

The TRS R consists of the following rules:

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


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

POL(A(x1)) = 4 + 2·x1   
POL(B(x1)) = 2·x1   
POL(a(x1)) = 4 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 1   
POL(d(x1)) = 4 + 4·x1   

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

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

(8) Obligation:

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

A(x) → B(b(b(x)))

The TRS R consists of the following rules:

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

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(10) TRUE