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 QDPOrderProof (⇔)
↳8 QDP
↳9 DependencyGraphProof (⇔)
↳10 TRUE

(0) Obligation:

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

b(b(y, z), c(a, a, a)) → f(c(z, y, z))
f(b(b(a, z), c(a, x, y))) → z
c(y, x, f(z)) → b(f(b(z, x)), z)

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:

B(b(y, z), c(a, a, a)) → F(c(z, y, z))
B(b(y, z), c(a, a, a)) → C(z, y, z)
C(y, x, f(z)) → B(f(b(z, x)), z)
C(y, x, f(z)) → F(b(z, x))
C(y, x, f(z)) → B(z, x)

The TRS R consists of the following rules:

b(b(y, z), c(a, a, a)) → f(c(z, y, z))
f(b(b(a, z), c(a, x, y))) → z
c(y, x, f(z)) → b(f(b(z, x)), z)

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 2 less nodes.

(4) Obligation:

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

B(b(y, z), c(a, a, a)) → C(z, y, z)
C(y, x, f(z)) → B(f(b(z, x)), z)
C(y, x, f(z)) → B(z, x)

The TRS R consists of the following rules:

b(b(y, z), c(a, a, a)) → f(c(z, y, z))
f(b(b(a, z), c(a, x, y))) → z
c(y, x, f(z)) → b(f(b(z, x)), z)

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(y, x, f(z)) → B(z, x)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(B(x1, x2)) = 0A + 0A·x1 + 0A·x2

POL(b(x1, x2)) = 0A + 1A·x1 + 1A·x2

POL(c(x1, x2, x3)) = 1A + -I·x1 + 2A·x2 + 2A·x3

POL(a) = 0A

POL(C(x1, x2, x3)) = 2A + -I·x1 + 1A·x2 + 1A·x3

POL(f(x1)) = -I + 0A·x1

The following usable rules [FROCOS05] were oriented:

b(b(y, z), c(a, a, a)) → f(c(z, y, z))
c(y, x, f(z)) → b(f(b(z, x)), z)
f(b(b(a, z), c(a, x, y))) → z

(6) Obligation:

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

B(b(y, z), c(a, a, a)) → C(z, y, z)
C(y, x, f(z)) → B(f(b(z, x)), z)

The TRS R consists of the following rules:

b(b(y, z), c(a, a, a)) → f(c(z, y, z))
f(b(b(a, z), c(a, x, y))) → z
c(y, x, f(z)) → b(f(b(z, x)), z)

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.


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

POL(B(x1, x2)) =
/0\
\0/
 +
/10\
\10/
·x1 +
/01\
\01/
·x2

POL(b(x1, x2)) =
/0\
\0/
 +
/10\
\10/
·x1 +
/13\
\00/
·x2

POL(c(x1, x2, x3)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2 +
/12\
\01/
·x3

POL(a) =
/0\
\2/

POL(C(x1, x2, x3)) =
/0\
\2/
 +
/01\
\10/
·x1 +
/00\
\10/
·x2 +
/02\
\01/
·x3

POL(f(x1)) =
/0\
\0/
 +
/01\
\11/
·x1

The following usable rules [FROCOS05] were oriented:

b(b(y, z), c(a, a, a)) → f(c(z, y, z))
c(y, x, f(z)) → b(f(b(z, x)), z)
f(b(b(a, z), c(a, x, y))) → z

(8) Obligation:

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

C(y, x, f(z)) → B(f(b(z, x)), z)

The TRS R consists of the following rules:

b(b(y, z), c(a, a, a)) → f(c(z, y, z))
f(b(b(a, z), c(a, x, y))) → z
c(y, x, f(z)) → b(f(b(z, x)), z)

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