YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_06/13.srs

(0) Obligation:

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

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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

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

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


(2) Obligation:

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

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

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

The TRS R consists of the following rules:

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


Used ordering: Polynomial interpretation [POLO]:

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

(6) Obligation:

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

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

The TRS R consists of the following rules:

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


A(a(a(b(b(x))))) → B(b(b(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,0A,0A]
·x1

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

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

POL(B(x1)) = 0A +
[0A,-I,-I]
·x1

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

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

(8) Obligation:

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

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

The TRS R consists of the following rules:

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