Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/Ste92SLASHhydra.trs

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) = x2
f₁(x1) = x1
nil = nil
copy₃(x1, x2, x3) = x1
n = n

Lexicographic Path Order [19].
Precedence:

COPY₁ > n
s₁ > n
nil > 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.