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:

g2(f2(x, y), z) -> f2(x, g2(y, z))
g2(h2(x, y), z) -> g2(x, f2(y, z))
g2(x, h2(y, z)) -> h2(g2(x, y), z)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

g2(f2(x, y), z) -> f2(x, g2(y, z))
g2(h2(x, y), z) -> g2(x, f2(y, z))
g2(x, h2(y, z)) -> h2(g2(x, y), z)

Q is empty.

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

G2(h2(x, y), z) -> G2(x, f2(y, z))
G2(f2(x, y), z) -> G2(y, z)
G2(x, h2(y, z)) -> G2(x, y)

The TRS R consists of the following rules:

g2(f2(x, y), z) -> f2(x, g2(y, z))
g2(h2(x, y), z) -> g2(x, f2(y, z))
g2(x, h2(y, z)) -> h2(g2(x, y), z)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof

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

G2(h2(x, y), z) -> G2(x, f2(y, z))
G2(f2(x, y), z) -> G2(y, z)
G2(x, h2(y, z)) -> G2(x, y)

The TRS R consists of the following rules:

g2(f2(x, y), z) -> f2(x, g2(y, z))
g2(h2(x, y), z) -> g2(x, f2(y, z))
g2(x, h2(y, z)) -> h2(g2(x, y), z)

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, h2(y, z)) -> G2(x, y)
Used argument filtering: G2(x1, x2)  =  x2
f2(x1, x2)  =  f
h2(x1, x2)  =  h1(x1)
Used ordering: Quasi Precedence: h_1 > f 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ QDPAfsSolverProof

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

G2(h2(x, y), z) -> G2(x, f2(y, z))
G2(f2(x, y), z) -> G2(y, z)

The TRS R consists of the following rules:

g2(f2(x, y), z) -> f2(x, g2(y, z))
g2(h2(x, y), z) -> g2(x, f2(y, z))
g2(x, h2(y, z)) -> h2(g2(x, y), z)

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(f2(x, y), z) -> G2(y, z)
Used argument filtering: G2(x1, x2)  =  x1
h2(x1, x2)  =  x1
f2(x1, x2)  =  f1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ QDPAfsSolverProof

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

G2(h2(x, y), z) -> G2(x, f2(y, z))

The TRS R consists of the following rules:

g2(f2(x, y), z) -> f2(x, g2(y, z))
g2(h2(x, y), z) -> g2(x, f2(y, z))
g2(x, h2(y, z)) -> h2(g2(x, y), z)

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(h2(x, y), z) -> G2(x, f2(y, z))
Used argument filtering: G2(x1, x2)  =  x1
h2(x1, x2)  =  h1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ QDPAfsSolverProof
                ↳ QDP
                  ↳ PisEmptyProof

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

g2(f2(x, y), z) -> f2(x, g2(y, z))
g2(h2(x, y), z) -> g2(x, f2(y, z))
g2(x, h2(y, z)) -> h2(g2(x, y), z)

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.