Termination w.r.t. Q proof of TRS/nontermin/AG01#4.34.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, x, y) -> x
if₃(false, x, y) -> y
g₂(s₁(x), s₁(y)) -> if₃(f₁(x), s₁(x), s₁(y))
g₂(x, c₁(y)) -> g₂(x, g₂(s₁(c₁(y)), 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, x, y) -> x
if₃(false, x, y) -> y
g₂(s₁(x), s₁(y)) -> if₃(f₁(x), s₁(x), s₁(y))
g₂(x, c₁(y)) -> g₂(x, g₂(s₁(c₁(y)), 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, g₂(s₁(c₁(y)), y))
G₂(s₁(x), s₁(y)) -> IF₃(f₁(x), s₁(x), s₁(y))
G₂(s₁(x), s₁(y)) -> F₁(x)
G₂(x, c₁(y)) -> G₂(s₁(c₁(y)), y)
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, x, y) -> x
if₃(false, x, y) -> y
g₂(s₁(x), s₁(y)) -> if₃(f₁(x), s₁(x), s₁(y))
g₂(x, c₁(y)) -> g₂(x, g₂(s₁(c₁(y)), 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, g₂(s₁(c₁(y)), y))
G₂(s₁(x), s₁(y)) -> IF₃(f₁(x), s₁(x), s₁(y))
G₂(s₁(x), s₁(y)) -> F₁(x)
G₂(x, c₁(y)) -> G₂(s₁(c₁(y)), y)
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, x, y) -> x
if₃(false, x, y) -> y
g₂(s₁(x), s₁(y)) -> if₃(f₁(x), s₁(x), s₁(y))
g₂(x, c₁(y)) -> g₂(x, g₂(s₁(c₁(y)), 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 2 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, x, y) -> x
if₃(false, x, y) -> y
g₂(s₁(x), s₁(y)) -> if₃(f₁(x), s₁(x), s₁(y))
g₂(x, c₁(y)) -> g₂(x, g₂(s₁(c₁(y)), 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 oriented strictly and are deleted.

F₁(s₁(x)) -> F₁(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F₁(x₁)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

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

The TRS R consists of the following rules:

f₁(0) -> true
f₁(1) -> false
f₁(s₁(x)) -> f₁(x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
g₂(s₁(x), s₁(y)) -> if₃(f₁(x), s₁(x), s₁(y))
g₂(x, c₁(y)) -> g₂(x, g₂(s₁(c₁(y)), 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 oriented strictly and are deleted.

G₂(x, c₁(y)) -> G₂(s₁(c₁(y)), y)

The remaining pairs can at least be oriented weakly.

G₂(x, c₁(y)) -> G₂(x, g₂(s₁(c₁(y)), y))

Used ordering: Polynomial interpretation [21]:

POL(0) = 2
POL(1) = 2
POL(G₂(x₁, x₂)) = x₁ + 2·x₂
POL(c₁(x₁)) = 2 + 2·x₁
POL(f₁(x₁)) = 2 + 2·x₁
POL(false) = 1
POL(g₂(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(if₃(x₁, x₂, x₃)) = 2·x₂ + 2·x₃
POL(s₁(x₁)) = 1
POL(true) = 2

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

G₂(x, c₁(y)) -> G₂(x, g₂(s₁(c₁(y)), y))

The TRS R consists of the following rules:

f₁(0) -> true
f₁(1) -> false
f₁(s₁(x)) -> f₁(x)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
g₂(s₁(x), s₁(y)) -> if₃(f₁(x), s₁(x), s₁(y))
g₂(x, c₁(y)) -> g₂(x, g₂(s₁(c₁(y)), 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 oriented strictly and are deleted.

G₂(x, c₁(y)) -> G₂(x, g₂(s₁(c₁(y)), y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 2
POL(1) = 2
POL(G₂(x₁, x₂)) = x₂
POL(c₁(x₁)) = 2 + x₁
POL(f₁(x₁)) = 2 + 2·x₁
POL(false) = 1
POL(g₂(x₁, x₂)) = 1 + 2·x₁
POL(if₃(x₁, x₂, x₃)) = 2·x₂ + 2·x₃
POL(s₁(x₁)) = 0
POL(true) = 2

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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, x, y) -> x
if₃(false, x, y) -> y
g₂(s₁(x), s₁(y)) -> if₃(f₁(x), s₁(x), s₁(y))
g₂(x, c₁(y)) -> g₂(x, g₂(s₁(c₁(y)), 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.