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-10.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 QDP
↳7 QDPOrderProof (⇔, 122 ms)
↳8 QDP
↳9 PisEmptyProof (⇔, 0 ms)
↳10 YES

(0) Obligation:

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

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

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

The TRS R consists of the following rules:

a(x) → x
a(x) → b(x)
c(b(a(x))) → a(b(a(c(c(x)))))
c(x) → 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 2 less nodes.

(6) Obligation:

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

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

The TRS R consists of the following rules:

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


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

POL(C(x1)) = 0A +
[0A,-I,-I]
·x1

POL(b(x1)) =
/1A\
|-I|
\-I/
 +
/-I0A0A\
|-I-I-I|
\-I-I-I/
·x1

POL(a(x1)) =
/1A\
|-I|
\0A/
 +
/0A0A0A\
|1A0A-I|
\-I0A0A/
·x1

POL(c(x1)) =
/-I\
|-I|
\-I/
 +
/0A-I-I\
|1A0A-I|
\0A0A0A/
·x1

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

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

(8) Obligation:

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

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