YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Mixed_SRS/03.srs

(0) Obligation:

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

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

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

B(a(a(a(x)))) → B(b(x))
B(a(a(a(x)))) → B(x)
B(b(x)) → B(a(x))

The TRS R consists of the following rules:

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

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

(5) 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:

B(a(a(a(x)))) → B(x)
B(b(x)) → B(a(x))


Used ordering: Polynomial interpretation [POLO]:

POL(B(x1)) = x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 3 + x1   

(6) Obligation:

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

B(a(a(a(x)))) → B(b(x))

The TRS R consists of the following rules:

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


B(a(a(a(x)))) → B(b(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic integers [ARCTIC,STERNAGEL_THIEMANN_RTA14]:

POL(B(x1)) = -I +
[0A,0A,-I]
·x1

POL(a(x1)) =
/0A\
|-1A|
\-I/
+
/-I0A-I\
|2A-I0A|
\-1A-I-I/
·x1

POL(b(x1)) =
/-I\
|-I|
\1A/
+
/0A-1A1A\
|0A-1A1A|
\2A1A-1A/
·x1

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

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

(8) Obligation:

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

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