Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


B(z, b(c(a, y, a), f(f(x)))) → C(y, a, z)
B(z, b(c(a, y, a), f(f(x)))) → C(c(y, a, z), z, x)
The remaining pairs can at least be oriented weakly.

C(c(z, y, a), a, a) → B(z, y)
Used ordering: Polynomial interpretation [25]:

POL(B(x1, x2)) = x1 + x2   
POL(C(x1, x2, x3)) = x1   
POL(a) = 1   
POL(b(x1, x2)) = x1 + x2   
POL(c(x1, x2, x3)) = x1 + x2   
POL(f(x1)) = 1   

The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof
          ↳ QDP

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

C(c(z, y, a), a, a) → B(z, y)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

F(c(x, y, z)) → F(b(y, z))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(c(x, y, z)) → F(b(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( a ) =
/0\
\0/

M( f(x1) ) =
/0\
\1/
+
/01\
\11/
·x1

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

M( c(x1, ..., x3) ) =
/0\
\1/
+
/10\
\10/
·x1+
/10\
\00/
·x2+
/00\
\10/
·x3

Tuple symbols:
M( F(x1) ) = 0+
[0,1]
·x1


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.