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.

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 contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

COPY3(s1(x), y, z) -> COPY3(x, y, cons2(f1(y), z))
Used argument filtering: COPY3(x1, x2, x3)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


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