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(cons2(nil, y)) -> y
f1(cons2(f1(cons2(nil, y)), z)) -> copy3(n, y, z)
copy3(0, y, z) -> f1(z)
copy3(s1(x), y, z) -> copy3(x, y, cons2(f1(y), z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(cons2(nil, y)) -> y
f1(cons2(f1(cons2(nil, y)), z)) -> copy3(n, y, z)
copy3(0, y, z) -> f1(z)
copy3(s1(x), y, z) -> copy3(x, y, cons2(f1(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:

COPY3(s1(x), y, z) -> F1(y)
COPY3(0, y, z) -> F1(z)
COPY3(s1(x), y, z) -> COPY3(x, y, cons2(f1(y), z))
F1(cons2(f1(cons2(nil, y)), z)) -> COPY3(n, y, z)

The TRS R consists of the following rules:

f1(cons2(nil, y)) -> y
f1(cons2(f1(cons2(nil, y)), z)) -> copy3(n, y, z)
copy3(0, y, z) -> f1(z)
copy3(s1(x), y, z) -> copy3(x, y, cons2(f1(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:

COPY3(s1(x), y, z) -> F1(y)
COPY3(0, y, z) -> F1(z)
COPY3(s1(x), y, z) -> COPY3(x, y, cons2(f1(y), z))
F1(cons2(f1(cons2(nil, y)), z)) -> COPY3(n, y, z)

The TRS R consists of the following rules:

f1(cons2(nil, y)) -> y
f1(cons2(f1(cons2(nil, y)), z)) -> copy3(n, y, z)
copy3(0, y, z) -> f1(z)
copy3(s1(x), y, z) -> copy3(x, y, cons2(f1(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 1 SCC with 3 less nodes.

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

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

COPY3(s1(x), y, z) -> COPY3(x, y, cons2(f1(y), z))

The TRS R consists of the following rules:

f1(cons2(nil, y)) -> y
f1(cons2(f1(cons2(nil, y)), z)) -> copy3(n, y, z)
copy3(0, y, z) -> f1(z)
copy3(s1(x), y, z) -> copy3(x, y, cons2(f1(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.


COPY3(s1(x), y, z) -> COPY3(x, y, cons2(f1(y), z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(COPY3(x1, x2, x3)) = 2·x1 + x1·x2 + 3·x1·x2·x3 + 3·x1·x3 + 2·x2·x3 + 3·x3   
POL(cons2(x1, x2)) = x2   
POL(copy3(x1, x2, x3)) = 0   
POL(f1(x1)) = 0   
POL(n) = 0   
POL(nil) = 0   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



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

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

f1(cons2(nil, y)) -> y
f1(cons2(f1(cons2(nil, y)), z)) -> copy3(n, y, z)
copy3(0, y, z) -> f1(z)
copy3(s1(x), y, z) -> copy3(x, y, cons2(f1(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.