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.

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 contains 2 SCCs.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F3(s1(x), y, z) -> F3(x, s1(c1(y)), c1(z))
Used argument filtering: F3(x1, x2, x3)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
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.