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-549.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 MRRProof (⇔, 36 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 67 ms)
↳6 QDP
↳7 QDPOrderProof (⇔, 51 ms)
↳8 QDP
↳9 DependencyGraphProof (⇔, 0 ms)
↳10 TRUE

(0) Obligation:

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

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

The TRS R consists of the following rules:

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


Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = x1   
POL(C(x1)) = 2 + 2·x1   
POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(c(x1)) = 2 + 2·x1   

(4) Obligation:

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

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

The TRS R consists of the following rules:

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


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

POL(A(x1)) = 1A +
[0A,0A,0A]
·x1

POL(c(x1)) =
/0A\
|0A|
\-I/
 +
/0A0A-I\
|0A0A1A|
\-I-I0A/
·x1

POL(C(x1)) = 0A +
[0A,0A,1A]
·x1

POL(a(x1)) =
/0A\
|-I|
\-I/
 +
/-I0A0A\
|-I-I0A|
\-I0A-I/
·x1

POL(b(x1)) =
/0A\
|-I|
\0A/
 +
/0A0A1A\
|-I-I0A|
\0A0A1A/
·x1

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

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

(6) Obligation:

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

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

The TRS R consists of the following rules:

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

POL(A(x1)) = -I +
[0A,-I,0A]
·x1

POL(c(x1)) =
/1A\
|-I|
\0A/
 +
/0A1A1A\
|-I-I-I|
\-I0A0A/
·x1

POL(C(x1)) = 0A +
[-I,0A,0A]
·x1

POL(a(x1)) =
/0A\
|-I|
\0A/
 +
/-I0A0A\
|-I0A-I|
\0A0A-I/
·x1

POL(b(x1)) =
/0A\
|-I|
\1A/
 +
/-I0A0A\
|-I1A-I|
\0A0A1A/
·x1

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

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

(8) Obligation:

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

C(b(x)) → A(c(x))

The TRS R consists of the following rules:

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