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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 Narrowing (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 QDP
↳9 Narrowing (⇔)
↳10 QDP
↳11 QDPOrderProof (⇔)
↳12 QDP
↳13 QDPOrderProof (⇔)
↳14 QDP
↳15 Narrowing (⇔)
↳16 QDP
↳17 UsableRulesProof (⇔)
↳18 QDP
↳19 MNOCProof (⇔)
↳20 QDP
↳21 Rewriting (⇔)
↳22 QDP
↳23 DependencyGraphProof (⇔)
↳24 TRUE

(0) Obligation:

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

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

(5) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule C(a(a(0, x), y)) → C(c(0)) at position [0] we obtained the following new rules [LPAR04]:

C(a(a(0, y0), y1)) → C(0)

(6) Obligation:

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

C(c(c(y))) → C(a(y, 0))
C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(y))) → C(c(a(y, 0)))
C(a(a(0, y0), y1)) → C(0)

The TRS R consists of the following rules:

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

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 1 SCC with 1 less node.

(8) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(9) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule C(c(c(y))) → C(c(a(y, 0))) at position [0] we obtained the following new rules [LPAR04]:

C(c(c(a(0, x0)))) → C(a(c(c(c(0))), 0))
C(c(c(y0))) → C(a(y0, 0))

(10) Obligation:

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

C(a(a(0, x), y)) → C(c(c(0)))
C(c(c(y))) → C(a(y, 0))
C(c(c(a(0, x0)))) → C(a(c(c(c(0))), 0))

The TRS R consists of the following rules:

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

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


C(c(c(a(0, x0)))) → C(a(c(c(c(0))), 0))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(C(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

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

POL(0) =
/0\
\0/

POL(c(x1)) =
/1\
\0/
 +
/10\
\11/
·x1

The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

POL(C(x1)) =
/0\
\0/
 +
/20\
\00/
·x1

POL(a(x1, x2)) =
/0\
\2/
 +
/02\
\00/
·x1 +
/00\
\00/
·x2

POL(0) =
/0\
\0/

POL(c(x1)) =
/0\
\2/
 +
/22\
\01/
·x1

The following usable rules [FROCOS05] were oriented:

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

(14) Obligation:

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

C(a(a(0, x), y)) → C(c(c(0)))

The TRS R consists of the following rules:

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

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

(15) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule C(a(a(0, x), y)) → C(c(c(0))) at position [0] we obtained the following new rules [LPAR04]:

C(a(a(0, y0), y1)) → C(c(0))

(16) Obligation:

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

C(a(a(0, y0), y1)) → C(c(0))

The TRS R consists of the following rules:

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

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

(17) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(18) Obligation:

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

C(a(a(0, y0), y1)) → C(c(0))

The TRS R consists of the following rules:

c(y) → y

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

(19) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(20) Obligation:

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

C(a(a(0, y0), y1)) → C(c(0))

The TRS R consists of the following rules:

c(y) → y

The set Q consists of the following terms:

c(x0)

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

(21) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule C(a(a(0, y0), y1)) → C(c(0)) at position [0] we obtained the following new rules [LPAR04]:

C(a(a(0, y0), y1)) → C(0)

(22) Obligation:

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

C(a(a(0, y0), y1)) → C(0)

The TRS R consists of the following rules:

c(y) → y

The set Q consists of the following terms:

c(x0)

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

(23) DependencyGraphProof (EQUIVALENT transformation)

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

(24) TRUE