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

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

(0) Obligation:

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

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

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

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

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

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

POL(C(x1)) =
/0\
\1/
 +
/10\
\00/
·x1

POL(c(x1)) =
/0\
\0/
 +
/11\
\10/
·x1

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

POL(b(x1, x2)) =
/1\
\1/
 +
/10\
\00/
·x1 +
/00\
\01/
·x2

POL(0) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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

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

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

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) TRUE