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

(0) Obligation:

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

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

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

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

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

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(f(z), f(c(a, x, a)), y) → B(x, z)

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(B(x1, x2)) =

/ 0A \ | -I | \ 3A /

+

/ 0A 0A 0A \ | -I -I -I | \ 0A 1A 1A /

·

x1

+

/ 0A 0A 0A \ | -I -I -I | \ 1A 1A 1A /

·

x2

POL(C(x1, x2, x3)) =

/ 0A \ | -I | \ 2A /

+

/ 0A 0A 0A \ | -I -I -I | \ 0A 1A 1A /

·

x1

+

/ 0A 0A -I \ | -I -I -I | \ 0A 1A -I /

·

x2

+

/ 0A 0A 0A \ | -I -I -I | \ 1A 1A 1A /

·

x3

POL(f(x1)) =

/ -I \ | 0A | \ -I /

+

/ 0A 0A -I \ | 0A -I 1A | \ 1A 1A -I /

·

x1

POL(c(x1, x2, x3)) =

/ 0A \ | -I | \ 0A /

+

/ -I -I -I \ | 1A -I -I | \ 1A -I -I /

·

x1

+

/ -I -I -I \ | 0A 0A 0A | \ 0A 0A 0A /

·

x2

+

/ -I -I -I \ | 0A 0A 0A | \ 0A 0A 0A /

·

x3

POL(a) =

/ 1A \ | 0A | \ 0A /

POL(b(x1, x2)) =

/ 2A \ | 2A | \ 2A /

+

/ 0A 0A 0A \ | 0A 0A 0A | \ -I 0A 0A /

·

x1

+

/ -I -I -I \ | -I 0A -I | \ -I 0A 0A /

·

x2

The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

(9) Instantiation (EQUIVALENT transformation)

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

C(f(x0), f(c(a, x1, a)), a) → C(x0, a, a)

(10) Obligation:

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

C(f(z), f(c(a, x, a)), y) → C(f(b(x, z)), c(z, y, a), a)
C(f(x0), f(c(a, x1, a)), a) → C(x0, a, a)

The TRS R consists of the following rules:

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

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

(11) DependencyGraphProof (EQUIVALENT transformation)

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

(12) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(13) Instantiation (EQUIVALENT transformation)

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

C(f(y_2), f(c(a, x1, a)), a) → C(f(b(x1, y_2)), c(y_2, a, a), a)

(14) Obligation:

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

C(f(y_2), f(c(a, x1, a)), a) → C(f(b(x1, y_2)), c(y_2, a, a), a)

The TRS R consists of the following rules:

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

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

(15) DependencyGraphProof (EQUIVALENT transformation)

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

(16) TRUE