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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 Instantiation (⇔)
↳6 QDP
↳7 Instantiation (⇔)
↳8 QDP
↳9 DependencyGraphProof (⇔)
↳10 QDP
↳11 QDPOrderProof (⇔)
↳12 QDP
↳13 DependencyGraphProof (⇔)
↳14 QDP
↳15 QDPSizeChangeProof (⇔)
↳16 TRUE

(0) Obligation:

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

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

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

The TRS R consists of the following rules:

c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → 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 4 less nodes.

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(5) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule C(z, x, a) → B(b(f(z), z), x) we obtained the following new rules [LPAR04]:

C(f(a), y_0, a) → B(b(f(f(a)), f(a)), y_0)

(6) Obligation:

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

B(y, b(z, a)) → B(c(f(a), y, z), z)
B(y, b(z, a)) → C(f(a), y, z)
C(z, x, a) → B(f(z), z)
C(f(a), y_0, a) → B(b(f(f(a)), f(a)), y_0)

The TRS R consists of the following rules:

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

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

(7) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule C(z, x, a) → B(f(z), z) we obtained the following new rules [LPAR04]:

C(f(a), y_0, a) → B(f(f(a)), f(a))

(8) Obligation:

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

B(y, b(z, a)) → B(c(f(a), y, z), z)
B(y, b(z, a)) → C(f(a), y, z)
C(f(a), y_0, a) → B(b(f(f(a)), f(a)), y_0)
C(f(a), y_0, a) → B(f(f(a)), f(a))

The TRS R consists of the following rules:

c(z, x, a) → f(b(b(f(z), z), x))
b(y, b(z, a)) → f(b(c(f(a), y, z), z))
f(c(c(z, a, a), x, a)) → 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 1 SCC with 1 less node.

(10) Obligation:

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

B(y, b(z, a)) → C(f(a), y, z)
C(f(a), y_0, a) → B(b(f(f(a)), f(a)), y_0)
B(y, b(z, a)) → B(c(f(a), y, z), z)

The TRS R consists of the following rules:

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

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.


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

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

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

POL(a) =
/0\
\1/

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

POL(f(x1)) =
/0\
\0/
 +
/10\
\10/
·x1

POL(c(x1, x2, x3)) =
/1\
\0/
 +
/11\
\11/
·x1 +
/11\
\00/
·x2 +
/00\
\00/
·x3

The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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

C(f(a), y_0, a) → B(b(f(f(a)), f(a)), y_0)
B(y, b(z, a)) → B(c(f(a), y, z), z)

The TRS R consists of the following rules:

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

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

(13) DependencyGraphProof (EQUIVALENT transformation)

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

(14) Obligation:

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

B(y, b(z, a)) → B(c(f(a), y, z), z)

The TRS R consists of the following rules:

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

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

(15) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • B(y, b(z, a)) → B(c(f(a), y, z), z)
    The graph contains the following edges 2 > 2

(16) TRUE