Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


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

B(y, b(a, z)) → B(f(c(y, y, a)), b(f(z), a))
Used ordering: Combined order from the following AFS and order.
B(x1, x2)  =  x2
b(x1, x2)  =  b(x2)
a  =  a
f(x1)  =  x1
c(x1, x2, x3)  =  c(x1, x2, x3)
F(x1)  =  F(x1)

Recursive path order with status [2].
Precedence:
b1 > c3 > a
b1 > F1 > a

Status:
a: multiset
b1: multiset
c3: multiset
F1: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP

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

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

The TRS R consists of the following rules:

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

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