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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 MRRProof (⇔)
↳6 QDP
↳7 QDPOrderProof (⇔)
↳8 QDP
↳9 DependencyGraphProof (⇔)
↳10 TRUE

(0) Obligation:

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

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(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:

C(c(b(c(x)))) → A(0, c(x))
C(c(x)) → C(b(c(x)))
A(0, x) → C(c(x))
A(0, x) → C(x)

The TRS R consists of the following rules:

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(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 1 less node.

(4) Obligation:

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

A(0, x) → C(c(x))
C(c(b(c(x)))) → A(0, c(x))
A(0, x) → C(x)

The TRS R consists of the following rules:

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(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(0, x) → C(x)


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(A(x1, x2)) = 1 + x1 + x2   
POL(C(x1)) = x1   
POL(a(x1, x2)) = 2 + x1 + x2   
POL(b(x1)) = x1   
POL(c(x1)) = 1 + x1   

(6) Obligation:

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

A(0, x) → C(c(x))
C(c(b(c(x)))) → A(0, c(x))

The TRS R consists of the following rules:

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(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].


The following pairs can be oriented strictly and are deleted.


C(c(b(c(x)))) → A(0, c(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(A(x1, x2)) =
/0\
\0/
 +
/11\
\00/
·x1 +
/01\
\00/
·x2

POL(0) =
/1\
\1/

POL(C(x1)) =
/1\
\0/
 +
/10\
\00/
·x1

POL(c(x1)) =
/1\
\0/
 +
/01\
\10/
·x1

POL(b(x1)) =
/0\
\1/
 +
/00\
\01/
·x1

POL(a(x1, x2)) =
/0\
\1/
 +
/11\
\00/
·x1 +
/10\
\01/
·x2

The following usable rules [FROCOS05] were oriented:

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(c(x))

(8) Obligation:

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

A(0, x) → C(c(x))

The TRS R consists of the following rules:

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(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