Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/SK90SLASH4.57.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:

f₃(x, y, z) -> g₄(<=₂(x, y), x, y, z)
g₄(true, x, y, z) -> z
g₄(false, x, y, z) -> f₃(f₃(p₁(x), y, z), f₃(p₁(y), z, x), f₃(p₁(z), x, y))
p₁(0) -> 0
p₁(s₁(x)) -> x

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

f₃(x, y, z) -> g₄(<=₂(x, y), x, y, z)
g₄(true, x, y, z) -> z
g₄(false, x, y, z) -> f₃(f₃(p₁(x), y, z), f₃(p₁(y), z, x), f₃(p₁(z), x, y))
p₁(0) -> 0
p₁(s₁(x)) -> x

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:

f₃(x, y, z) -> g₄(<=₂(x, y), x, y, z)
g₄(true, x, y, z) -> z
g₄(false, x, y, z) -> f₃(f₃(p₁(x), y, z), f₃(p₁(y), z, x), f₃(p₁(z), x, y))
p₁(0) -> 0
p₁(s₁(x)) -> x

The set Q consists of the following terms:

f₃(x0, x1, x2)
g₄(true, x0, x1, x2)
g₄(false, x0, x1, x2)
p₁(0)
p₁(s₁(x0))

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

G₄(false, x, y, z) -> P₁(z)
G₄(false, x, y, z) -> F₃(p₁(z), x, y)
G₄(false, x, y, z) -> F₃(p₁(x), y, z)
G₄(false, x, y, z) -> F₃(f₃(p₁(x), y, z), f₃(p₁(y), z, x), f₃(p₁(z), x, y))
G₄(false, x, y, z) -> P₁(y)
F₃(x, y, z) -> G₄(<=₂(x, y), x, y, z)
G₄(false, x, y, z) -> F₃(p₁(y), z, x)
G₄(false, x, y, z) -> P₁(x)

The TRS R consists of the following rules:

f₃(x, y, z) -> g₄(<=₂(x, y), x, y, z)
g₄(true, x, y, z) -> z
g₄(false, x, y, z) -> f₃(f₃(p₁(x), y, z), f₃(p₁(y), z, x), f₃(p₁(z), x, y))
p₁(0) -> 0
p₁(s₁(x)) -> x

The set Q consists of the following terms:

f₃(x0, x1, x2)
g₄(true, x0, x1, x2)
g₄(false, x0, x1, x2)
p₁(0)
p₁(s₁(x0))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

G₄(false, x, y, z) -> P₁(z)
G₄(false, x, y, z) -> F₃(p₁(z), x, y)
G₄(false, x, y, z) -> F₃(p₁(x), y, z)
G₄(false, x, y, z) -> F₃(f₃(p₁(x), y, z), f₃(p₁(y), z, x), f₃(p₁(z), x, y))
G₄(false, x, y, z) -> P₁(y)
F₃(x, y, z) -> G₄(<=₂(x, y), x, y, z)
G₄(false, x, y, z) -> F₃(p₁(y), z, x)
G₄(false, x, y, z) -> P₁(x)

The TRS R consists of the following rules:

f₃(x, y, z) -> g₄(<=₂(x, y), x, y, z)
g₄(true, x, y, z) -> z
g₄(false, x, y, z) -> f₃(f₃(p₁(x), y, z), f₃(p₁(y), z, x), f₃(p₁(z), x, y))
p₁(0) -> 0
p₁(s₁(x)) -> x

The set Q consists of the following terms:

f₃(x0, x1, x2)
g₄(true, x0, x1, x2)
g₄(false, x0, x1, x2)
p₁(0)
p₁(s₁(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 8 less nodes.