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 strictly oriented and are deleted.


COPY3(s1(x), y, z) -> COPY3(x, y, cons2(f1(y), z))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
COPY3(x1, x2, x3)  =  COPY1(x1)
s1(x1)  =  s1(x1)
cons2(x1, x2)  =  cons1(x2)
f1(x1)  =  f1(x1)
nil  =  nil
copy3(x1, x2, x3)  =  copy1(x1)
n  =  n

Lexicographic Path Order [19].
Precedence:
s1 > COPY1 > n
cons1 > copy1 > n
f1 > copy1 > n
nil > copy1 > n

The following usable rules [14] were oriented:

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



↳ 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.