YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Secret_06_SRS/secr10.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 13 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 15 ms)
↳6 QDP
↳7 QDPOrderProof (⇔, 103 ms)
↳8 QDP
↳9 DependencyGraphProof (⇔, 0 ms)
↳10 TRUE

(0) Obligation:

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

POL(A(x1)) = 1 + x1   
POL(C(x1)) = 1 + x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 0   
POL(c(x1)) = 1 + x1   

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

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

(6) Obligation:

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

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

The TRS R consists of the following rules:

a(a(x)) → c(b(a(b(a(x)))))
b(a(b(x))) → b(x)
a(a(a(x))) → c(c(a(x)))
c(c(x)) → a(b(c(b(a(x)))))
a(c(a(x))) → c(c(a(x)))
c(a(c(x))) → a(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(a(a(x))) → C(c(a(x)))
A(c(a(x))) → C(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,1A,-I]
·x1

POL(a(x1)) =
/-I\
|-I|
\0A/
 +
/0A1A0A\
|-I0A0A|
\0A1A0A/
·x1

POL(C(x1)) = 0A +
[-I,-I,0A]
·x1

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

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

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

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

(8) Obligation:

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

C(a(c(x))) → A(a(c(x)))

The TRS R consists of the following rules:

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