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:

f1(nil) -> nil
f1(.2(nil, y)) -> .2(nil, f1(y))
f1(.2(.2(x, y), z)) -> f1(.2(x, .2(y, z)))
g1(nil) -> nil
g1(.2(x, nil)) -> .2(g1(x), nil)
g1(.2(x, .2(y, z))) -> g1(.2(.2(x, y), z))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

f1(nil) -> nil
f1(.2(nil, y)) -> .2(nil, f1(y))
f1(.2(.2(x, y), z)) -> f1(.2(x, .2(y, z)))
g1(nil) -> nil
g1(.2(x, nil)) -> .2(g1(x), nil)
g1(.2(x, .2(y, z))) -> g1(.2(.2(x, y), z))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

f1(nil) -> nil
f1(.2(nil, y)) -> .2(nil, f1(y))
f1(.2(.2(x, y), z)) -> f1(.2(x, .2(y, z)))
g1(nil) -> nil
g1(.2(x, nil)) -> .2(g1(x), nil)
g1(.2(x, .2(y, z))) -> g1(.2(.2(x, y), z))

The set Q consists of the following terms:

f1(nil)
f1(.2(nil, x0))
f1(.2(.2(x0, x1), x2))
g1(nil)
g1(.2(x0, nil))
g1(.2(x0, .2(x1, x2)))


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

F1(.2(nil, y)) -> F1(y)
F1(.2(.2(x, y), z)) -> F1(.2(x, .2(y, z)))
G1(.2(x, .2(y, z))) -> G1(.2(.2(x, y), z))
G1(.2(x, nil)) -> G1(x)

The TRS R consists of the following rules:

f1(nil) -> nil
f1(.2(nil, y)) -> .2(nil, f1(y))
f1(.2(.2(x, y), z)) -> f1(.2(x, .2(y, z)))
g1(nil) -> nil
g1(.2(x, nil)) -> .2(g1(x), nil)
g1(.2(x, .2(y, z))) -> g1(.2(.2(x, y), z))

The set Q consists of the following terms:

f1(nil)
f1(.2(nil, x0))
f1(.2(.2(x0, x1), x2))
g1(nil)
g1(.2(x0, nil))
g1(.2(x0, .2(x1, x2)))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

F1(.2(nil, y)) -> F1(y)
F1(.2(.2(x, y), z)) -> F1(.2(x, .2(y, z)))
G1(.2(x, .2(y, z))) -> G1(.2(.2(x, y), z))
G1(.2(x, nil)) -> G1(x)

The TRS R consists of the following rules:

f1(nil) -> nil
f1(.2(nil, y)) -> .2(nil, f1(y))
f1(.2(.2(x, y), z)) -> f1(.2(x, .2(y, z)))
g1(nil) -> nil
g1(.2(x, nil)) -> .2(g1(x), nil)
g1(.2(x, .2(y, z))) -> g1(.2(.2(x, y), z))

The set Q consists of the following terms:

f1(nil)
f1(.2(nil, x0))
f1(.2(.2(x0, x1), x2))
g1(nil)
g1(.2(x0, nil))
g1(.2(x0, .2(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP

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

G1(.2(x, .2(y, z))) -> G1(.2(.2(x, y), z))
G1(.2(x, nil)) -> G1(x)

The TRS R consists of the following rules:

f1(nil) -> nil
f1(.2(nil, y)) -> .2(nil, f1(y))
f1(.2(.2(x, y), z)) -> f1(.2(x, .2(y, z)))
g1(nil) -> nil
g1(.2(x, nil)) -> .2(g1(x), nil)
g1(.2(x, .2(y, z))) -> g1(.2(.2(x, y), z))

The set Q consists of the following terms:

f1(nil)
f1(.2(nil, x0))
f1(.2(.2(x0, x1), x2))
g1(nil)
g1(.2(x0, nil))
g1(.2(x0, .2(x1, x2)))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof

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

F1(.2(nil, y)) -> F1(y)
F1(.2(.2(x, y), z)) -> F1(.2(x, .2(y, z)))

The TRS R consists of the following rules:

f1(nil) -> nil
f1(.2(nil, y)) -> .2(nil, f1(y))
f1(.2(.2(x, y), z)) -> f1(.2(x, .2(y, z)))
g1(nil) -> nil
g1(.2(x, nil)) -> .2(g1(x), nil)
g1(.2(x, .2(y, z))) -> g1(.2(.2(x, y), z))

The set Q consists of the following terms:

f1(nil)
f1(.2(nil, x0))
f1(.2(.2(x0, x1), x2))
g1(nil)
g1(.2(x0, nil))
g1(.2(x0, .2(x1, x2)))

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.

F1(.2(nil, y)) -> F1(y)
F1(.2(.2(x, y), z)) -> F1(.2(x, .2(y, z)))
Used argument filtering: F1(x1)  =  x1
.2(x1, x2)  =  .2(x1, x2)
nil  =  nil
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof

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

f1(nil) -> nil
f1(.2(nil, y)) -> .2(nil, f1(y))
f1(.2(.2(x, y), z)) -> f1(.2(x, .2(y, z)))
g1(nil) -> nil
g1(.2(x, nil)) -> .2(g1(x), nil)
g1(.2(x, .2(y, z))) -> g1(.2(.2(x, y), z))

The set Q consists of the following terms:

f1(nil)
f1(.2(nil, x0))
f1(.2(.2(x0, x1), x2))
g1(nil)
g1(.2(x0, nil))
g1(.2(x0, .2(x1, x2)))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.