Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

F₁(s₁(x)) -> F₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F₁(x1) = F₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > F₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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

            ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
G₂(x1, x2) = x2
c₁(x1) = c₁(x1)
s₁(x1) = s

Lexicographic Path Order [19].
Precedence:

[c₁, s]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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.