Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/SK90SLASH4.42.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:

f₂(a, g₁(y)) -> g₁(g₁(y))
f₂(g₁(x), a) -> f₂(x, g₁(a))
f₂(g₁(x), g₁(y)) -> h₃(g₁(y), x, g₁(y))
h₃(g₁(x), y, z) -> f₂(y, h₃(x, y, z))
h₃(a, y, z) -> z

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

f₂(a, g₁(y)) -> g₁(g₁(y))
f₂(g₁(x), a) -> f₂(x, g₁(a))
f₂(g₁(x), g₁(y)) -> h₃(g₁(y), x, g₁(y))
h₃(g₁(x), y, z) -> f₂(y, h₃(x, y, z))
h₃(a, y, z) -> z

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

f₂(a, g₁(y)) -> g₁(g₁(y))
f₂(g₁(x), a) -> f₂(x, g₁(a))
f₂(g₁(x), g₁(y)) -> h₃(g₁(y), x, g₁(y))
h₃(g₁(x), y, z) -> f₂(y, h₃(x, y, z))
h₃(a, y, z) -> z

The set Q consists of the following terms:

f₂(a, g₁(x0))
f₂(g₁(x0), a)
f₂(g₁(x0), g₁(x1))
h₃(g₁(x0), x1, x2)
h₃(a, x0, x1)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

H₃(g₁(x), y, z) -> H₃(x, y, z)
H₃(g₁(x), y, z) -> F₂(y, h₃(x, y, z))
F₂(g₁(x), g₁(y)) -> H₃(g₁(y), x, g₁(y))
F₂(g₁(x), a) -> F₂(x, g₁(a))

The TRS R consists of the following rules:

f₂(a, g₁(y)) -> g₁(g₁(y))
f₂(g₁(x), a) -> f₂(x, g₁(a))
f₂(g₁(x), g₁(y)) -> h₃(g₁(y), x, g₁(y))
h₃(g₁(x), y, z) -> f₂(y, h₃(x, y, z))
h₃(a, y, z) -> z

The set Q consists of the following terms:

f₂(a, g₁(x0))
f₂(g₁(x0), a)
f₂(g₁(x0), g₁(x1))
h₃(g₁(x0), x1, x2)
h₃(a, x0, x1)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

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

H₃(g₁(x), y, z) -> H₃(x, y, z)
H₃(g₁(x), y, z) -> F₂(y, h₃(x, y, z))
F₂(g₁(x), g₁(y)) -> H₃(g₁(y), x, g₁(y))
F₂(g₁(x), a) -> F₂(x, g₁(a))

The TRS R consists of the following rules:

f₂(a, g₁(y)) -> g₁(g₁(y))
f₂(g₁(x), a) -> f₂(x, g₁(a))
f₂(g₁(x), g₁(y)) -> h₃(g₁(y), x, g₁(y))
h₃(g₁(x), y, z) -> f₂(y, h₃(x, y, z))
h₃(a, y, z) -> z

The set Q consists of the following terms:

f₂(a, g₁(x0))
f₂(g₁(x0), a)
f₂(g₁(x0), g₁(x1))
h₃(g₁(x0), x1, x2)
h₃(a, x0, x1)

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.

F₂(g₁(x), a) -> F₂(x, g₁(a))

The remaining pairs can at least by weakly be oriented.

H₃(g₁(x), y, z) -> H₃(x, y, z)
H₃(g₁(x), y, z) -> F₂(y, h₃(x, y, z))
F₂(g₁(x), g₁(y)) -> H₃(g₁(y), x, g₁(y))

Used ordering: Combined order from the following AFS and order.
H₃(x1, x2, x3) = x3
g₁(x1) = g
F₂(x1, x2) = x2
h₃(x1, x2, x3) = x3
a = a
f₂(x1, x2) = x2

Lexicographic Path Order [19].
Precedence:

a > g

The following usable rules [14] were oriented:

f₂(g₁(x), g₁(y)) -> h₃(g₁(y), x, g₁(y))
h₃(g₁(x), y, z) -> f₂(y, h₃(x, y, z))
f₂(g₁(x), a) -> f₂(x, g₁(a))
h₃(a, y, z) -> z
f₂(a, g₁(y)) -> g₁(g₁(y))

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

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

F₂(g₁(x), g₁(y)) -> H₃(g₁(y), x, g₁(y))
H₃(g₁(x), y, z) -> F₂(y, h₃(x, y, z))
H₃(g₁(x), y, z) -> H₃(x, y, z)

The TRS R consists of the following rules:

f₂(a, g₁(y)) -> g₁(g₁(y))
f₂(g₁(x), a) -> f₂(x, g₁(a))
f₂(g₁(x), g₁(y)) -> h₃(g₁(y), x, g₁(y))
h₃(g₁(x), y, z) -> f₂(y, h₃(x, y, z))
h₃(a, y, z) -> z

The set Q consists of the following terms:

f₂(a, g₁(x0))
f₂(g₁(x0), a)
f₂(g₁(x0), g₁(x1))
h₃(g₁(x0), x1, x2)
h₃(a, x0, x1)

We have to consider all minimal (P,Q,R)-chains.