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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule B(x, b(z, y)) → F(b(f(f(z)), c(x, z, y))) we obtained the following new rules:

B(a, b(x1, x2)) → F(b(f(f(x1)), c(a, x1, x2)))



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

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

B(x, b(z, y)) → F(z)
F(c(a, z, x)) → B(a, z)
B(a, b(x1, x2)) → F(b(f(f(x1)), c(a, x1, x2)))
B(x, b(z, y)) → F(f(z))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule B(x, b(z, y)) → F(f(z)) we obtained the following new rules:

B(a, b(x1, x2)) → F(f(x1))



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

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

B(x, b(z, y)) → F(z)
F(c(a, z, x)) → B(a, z)
B(a, b(x1, x2)) → F(b(f(f(x1)), c(a, x1, x2)))
B(a, b(x1, x2)) → F(f(x1))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule B(x, b(z, y)) → F(z) we obtained the following new rules:

B(a, b(x1, x2)) → F(x1)



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

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

F(c(a, z, x)) → B(a, z)
B(a, b(x1, x2)) → F(x1)
B(a, b(x1, x2)) → F(b(f(f(x1)), c(a, x1, x2)))
B(a, b(x1, x2)) → F(f(x1))

The TRS R consists of the following rules:

f(c(a, z, x)) → b(a, z)
b(x, b(z, y)) → f(b(f(f(z)), c(x, z, y)))
b(y, z) → 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(a, b(x1, x2)) → F(x1)
B(a, b(x1, x2)) → F(b(f(f(x1)), c(a, x1, x2)))
B(a, b(x1, x2)) → F(f(x1))
The remaining pairs can at least be oriented weakly.

F(c(a, z, x)) → B(a, z)
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( f(x1) ) =
/1\
\0/
+
/01\
\01/
·x1

M( a ) =
/0\
\0/

M( b(x1, x2) ) =
/1\
\0/
+
/11\
\01/
·x1+
/11\
\11/
·x2

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

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

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



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

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

F(c(a, z, x)) → B(a, z)

The TRS R consists of the following rules:

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