YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-351.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 90 ms)
↳6 QDP
↳7 PisEmptyProof (⇔, 0 ms)
↳8 YES

(0) Obligation:

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

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

The TRS R consists of the following rules:

a(x) → b(x)
a(b(x)) → b(a(c(a(x))))
b(x) → c(x)
c(c(x)) → 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 4 less nodes.

(4) Obligation:

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

A(b(x)) → A(x)
A(b(x)) → A(c(a(x)))

The TRS R consists of the following rules:

a(x) → b(x)
a(b(x)) → b(a(c(a(x))))
b(x) → c(x)
c(c(x)) → 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(b(x)) → A(x)
A(b(x)) → A(c(a(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(A(x1)) = 0A +
[-I,0A,-I]
·x1

POL(b(x1)) =
/0A\
|1A|
\-I/
 +
/-I0A-I\
|0A1A-I|
\0A0A0A/
·x1

POL(c(x1)) =
/0A\
|-I|
\-I/
 +
/-I0A-I\
|0A-I-I|
\0A0A0A/
·x1

POL(a(x1)) =
/0A\
|1A|
\0A/
 +
/-I0A-I\
|0A1A-I|
\0A0A0A/
·x1

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

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

(6) Obligation:

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

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

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

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) YES