YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Secret_07_SRS/x03.srs

(0) Obligation:

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

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

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

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a(x1)) = 3 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 1 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

c(b(a(x))) → b(x)


(4) Obligation:

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

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

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(6) Obligation:

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

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

The TRS R consists of the following rules:

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

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

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


Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = 2 + x1   
POL(B(x1)) = x1   
POL(C(x1)) = x1   
POL(a(x1)) = 3 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 1 + x1   

(8) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


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

POL(B(x1)) = 0A +
[0A,0A,0A]
·x1

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

POL(c(x1)) =
/0A\
|0A|
\-I/
+
/-I-I0A\
|0A1A0A|
\-I0A-I/
·x1

POL(A(x1)) = 0A +
[0A,1A,0A]
·x1

POL(C(x1)) = 0A +
[0A,0A,0A]
·x1

POL(a(x1)) =
/0A\
|0A|
\-I/
+
/0A0A-I\
|0A1A-I|
\-I0A-I/
·x1

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

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

(10) Obligation:

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

B(b(c(x))) → A(x)
C(x) → B(x)

The TRS R consists of the following rules:

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

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(12) TRUE