Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/AG01SLASHHASH3.48.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(0) -> true
f₁(1) -> false
f₁(s₁(x)) -> f₁(x)
if₃(true, s₁(x), s₁(y)) -> s₁(x)
if₃(false, s₁(x), s₁(y)) -> s₁(y)
g₂(x, c₁(y)) -> c₁(g₂(x, y))
g₂(x, c₁(y)) -> g₂(x, if₃(f₁(x), c₁(g₂(s₁(x), y)), c₁(y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(0) -> true
f₁(1) -> false
f₁(s₁(x)) -> f₁(x)
if₃(true, s₁(x), s₁(y)) -> s₁(x)
if₃(false, s₁(x), s₁(y)) -> s₁(y)
g₂(x, c₁(y)) -> c₁(g₂(x, y))
g₂(x, c₁(y)) -> g₂(x, if₃(f₁(x), c₁(g₂(s₁(x), y)), c₁(y)))

Q is empty.

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

G₂(x, c₁(y)) -> G₂(x, if₃(f₁(x), c₁(g₂(s₁(x), y)), c₁(y)))
G₂(x, c₁(y)) -> G₂(x, y)
G₂(x, c₁(y)) -> IF₃(f₁(x), c₁(g₂(s₁(x), y)), c₁(y))
G₂(x, c₁(y)) -> G₂(s₁(x), y)
G₂(x, c₁(y)) -> F₁(x)
F₁(s₁(x)) -> F₁(x)

The TRS R consists of the following rules:

f₁(0) -> true
f₁(1) -> false
f₁(s₁(x)) -> f₁(x)
if₃(true, s₁(x), s₁(y)) -> s₁(x)
if₃(false, s₁(x), s₁(y)) -> s₁(y)
g₂(x, c₁(y)) -> c₁(g₂(x, y))
g₂(x, c₁(y)) -> g₂(x, if₃(f₁(x), c₁(g₂(s₁(x), y)), c₁(y)))

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:

G₂(x, c₁(y)) -> G₂(x, if₃(f₁(x), c₁(g₂(s₁(x), y)), c₁(y)))
G₂(x, c₁(y)) -> G₂(x, y)
G₂(x, c₁(y)) -> IF₃(f₁(x), c₁(g₂(s₁(x), y)), c₁(y))
G₂(x, c₁(y)) -> G₂(s₁(x), y)
G₂(x, c₁(y)) -> F₁(x)
F₁(s₁(x)) -> F₁(x)

The TRS R consists of the following rules:

f₁(0) -> true
f₁(1) -> false
f₁(s₁(x)) -> f₁(x)
if₃(true, s₁(x), s₁(y)) -> s₁(x)
if₃(false, s₁(x), s₁(y)) -> s₁(y)
g₂(x, c₁(y)) -> c₁(g₂(x, y))
g₂(x, c₁(y)) -> g₂(x, if₃(f₁(x), c₁(g₂(s₁(x), y)), c₁(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

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

F₁(s₁(x)) -> F₁(x)

The TRS R consists of the following rules:

f₁(0) -> true
f₁(1) -> false
f₁(s₁(x)) -> f₁(x)
if₃(true, s₁(x), s₁(y)) -> s₁(x)
if₃(false, s₁(x), s₁(y)) -> s₁(y)
g₂(x, c₁(y)) -> c₁(g₂(x, y))
g₂(x, c₁(y)) -> g₂(x, if₃(f₁(x), c₁(g₂(s₁(x), y)), c₁(y)))

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.

F₁(s₁(x)) -> F₁(x)

Used argument filtering: F₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

f₁(0) -> true
f₁(1) -> false
f₁(s₁(x)) -> f₁(x)
if₃(true, s₁(x), s₁(y)) -> s₁(x)
if₃(false, s₁(x), s₁(y)) -> s₁(y)
g₂(x, c₁(y)) -> c₁(g₂(x, y))
g₂(x, c₁(y)) -> g₂(x, if₃(f₁(x), c₁(g₂(s₁(x), y)), c₁(y)))

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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

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

G₂(x, c₁(y)) -> G₂(x, y)
G₂(x, c₁(y)) -> G₂(s₁(x), y)

The TRS R consists of the following rules:

f₁(0) -> true
f₁(1) -> false
f₁(s₁(x)) -> f₁(x)
if₃(true, s₁(x), s₁(y)) -> s₁(x)
if₃(false, s₁(x), s₁(y)) -> s₁(y)
g₂(x, c₁(y)) -> c₁(g₂(x, y))
g₂(x, c₁(y)) -> g₂(x, if₃(f₁(x), c₁(g₂(s₁(x), y)), c₁(y)))

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.

G₂(x, c₁(y)) -> G₂(x, y)
G₂(x, c₁(y)) -> G₂(s₁(x), y)

Used argument filtering: G₂(x1, x2) = x2
c₁(x1) = c₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

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

f₁(0) -> true
f₁(1) -> false
f₁(s₁(x)) -> f₁(x)
if₃(true, s₁(x), s₁(y)) -> s₁(x)
if₃(false, s₁(x), s₁(y)) -> s₁(y)
g₂(x, c₁(y)) -> c₁(g₂(x, y))
g₂(x, c₁(y)) -> g₂(x, if₃(f₁(x), c₁(g₂(s₁(x), y)), c₁(y)))

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.