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:

f3(x, c1(x), c1(y)) -> f3(y, y, f3(y, x, y))
f3(s1(x), y, z) -> f3(x, s1(c1(y)), c1(z))
f3(c1(x), x, y) -> c1(y)
g2(x, y) -> x
g2(x, y) -> y

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f3(x, c1(x), c1(y)) -> f3(y, y, f3(y, x, y))
f3(s1(x), y, z) -> f3(x, s1(c1(y)), c1(z))
f3(c1(x), x, y) -> c1(y)
g2(x, y) -> x
g2(x, y) -> y

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:

F3(s1(x), y, z) -> F3(x, s1(c1(y)), c1(z))
F3(x, c1(x), c1(y)) -> F3(y, y, f3(y, x, y))
F3(x, c1(x), c1(y)) -> F3(y, x, y)

The TRS R consists of the following rules:

f3(x, c1(x), c1(y)) -> f3(y, y, f3(y, x, y))
f3(s1(x), y, z) -> f3(x, s1(c1(y)), c1(z))
f3(c1(x), x, y) -> c1(y)
g2(x, y) -> x
g2(x, y) -> y

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:

F3(s1(x), y, z) -> F3(x, s1(c1(y)), c1(z))
F3(x, c1(x), c1(y)) -> F3(y, y, f3(y, x, y))
F3(x, c1(x), c1(y)) -> F3(y, x, y)

The TRS R consists of the following rules:

f3(x, c1(x), c1(y)) -> f3(y, y, f3(y, x, y))
f3(s1(x), y, z) -> f3(x, s1(c1(y)), c1(z))
f3(c1(x), x, y) -> c1(y)
g2(x, y) -> x
g2(x, y) -> y

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

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

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

F3(s1(x), y, z) -> F3(x, s1(c1(y)), c1(z))

The TRS R consists of the following rules:

f3(x, c1(x), c1(y)) -> f3(y, y, f3(y, x, y))
f3(s1(x), y, z) -> f3(x, s1(c1(y)), c1(z))
f3(c1(x), x, y) -> c1(y)
g2(x, y) -> x
g2(x, y) -> y

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.


F3(s1(x), y, z) -> F3(x, s1(c1(y)), c1(z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F3(x1, x2, x3)) = 2·x1 + x3   
POL(c1(x1)) = 0   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



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

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

f3(x, c1(x), c1(y)) -> f3(y, y, f3(y, x, y))
f3(s1(x), y, z) -> f3(x, s1(c1(y)), c1(z))
f3(c1(x), x, y) -> c1(y)
g2(x, y) -> x
g2(x, y) -> y

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.

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

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

F3(x, c1(x), c1(y)) -> F3(y, y, f3(y, x, y))
F3(x, c1(x), c1(y)) -> F3(y, x, y)

The TRS R consists of the following rules:

f3(x, c1(x), c1(y)) -> f3(y, y, f3(y, x, y))
f3(s1(x), y, z) -> f3(x, s1(c1(y)), c1(z))
f3(c1(x), x, y) -> c1(y)
g2(x, y) -> x
g2(x, y) -> y

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