Termination w.r.t. Q proof of TRS/Ste92hydra.trs

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(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

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.