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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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

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

The TRS R consists of the following rules:

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

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

Lexicographic path order with status [19].
Precedence:
a > b > F1
a > c > F1

Status:
b: multiset
a: multiset
c: multiset
F1: [1]

The following usable rules [14] were oriented:

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



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

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

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

The TRS R consists of the following rules:

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

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