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(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.


↳ QTRS
  ↳ DependencyPairsProof

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.

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

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

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

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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Instantiation

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

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

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.
By instantiating [15] the rule C(z, x, a) → B(b(f(z), z), x) we obtained the following new rules:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Instantiation
            ↳ QDP
              ↳ Instantiation

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

C(z, x, a) → B(f(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)
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.
By instantiating [15] the rule C(z, x, a) → B(f(z), z) we obtained the following new rules:

C(f(a), y_0, a) → B(f(f(a)), f(a))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Instantiation
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ DependencyGraphProof

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)
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.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Instantiation
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ QDPOrderProof

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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


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

C(f(a), y_0, a) → B(b(f(f(a)), f(a)), y_0)
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( a ) =
/1\
\0/

M( f(x1) ) =
/0\
\0/
+
/01\
\01/
·x1

M( b(x1, x2) ) =
/1\
\0/
+
/10\
\00/
·x1+
/10\
\00/
·x2

M( c(x1, ..., x3) ) =
/1\
\0/
+
/11\
\10/
·x1+
/00\
\00/
·x2+
/00\
\10/
·x3

Tuple symbols:
M( C(x1, ..., x3) ) = 0+
[0,0]
·x1+
[1,0]
·x2+
[1,0]
·x3

M( B(x1, x2) ) = 0+
[1,0]
·x1+
[1,0]
·x2


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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Instantiation
            ↳ QDP
              ↳ Instantiation
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ QDPOrderProof
                        ↳ QDP
                          ↳ DependencyGraphProof

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)

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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.