Termination w.r.t. Q of the following Term Rewriting System could be proven:

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
  ↳ 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))

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:

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))

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:

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))

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:

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))

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.


G1(.2(x, .2(y, z))) -> G1(.2(.2(x, y), z))
G1(.2(x, nil)) -> G1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(.2(x1, x2)) = 2 + x1 + 2·x2   
POL(G1(x1)) = 2·x1   
POL(nil) = 2   

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:

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.
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
            ↳ QDPOrderProof

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))

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.


F1(.2(nil, y)) -> F1(y)
F1(.2(.2(x, y), z)) -> F1(.2(x, .2(y, z)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(.2(x1, x2)) = 2 + 2·x1 + x2   
POL(F1(x1)) = 2·x1   
POL(nil) = 2   

The following usable rules [14] were oriented: none



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

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.