YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z114.srs

(0) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(3) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

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

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

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

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

Strictly oriented rules of the TRS R:

b(c(b(x))) → a(c(x))

Used ordering: Polynomial interpretation [POLO]:

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

(6) Obligation:

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

A(x) → B(b(b(x)))

The TRS R consists of the following rules:

a(c(x)) → c(b(x))
a(x) → b(b(b(x)))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(8) TRUE