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:

g(f(x, y), z) → f(x, g(y, z))
g(h(x, y), z) → g(x, f(y, z))
g(x, h(y, z)) → h(g(x, y), z)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

g(f(x, y), z) → f(x, g(y, z))
g(h(x, y), z) → g(x, f(y, z))
g(x, h(y, z)) → h(g(x, y), z)

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:

G(h(x, y), z) → G(x, f(y, z))
G(f(x, y), z) → G(y, z)
G(x, h(y, z)) → G(x, y)

The TRS R consists of the following rules:

g(f(x, y), z) → f(x, g(y, z))
g(h(x, y), z) → g(x, f(y, z))
g(x, h(y, z)) → h(g(x, y), z)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

G(h(x, y), z) → G(x, f(y, z))
G(f(x, y), z) → G(y, z)
G(x, h(y, z)) → G(x, y)

The TRS R consists of the following rules:

g(f(x, y), z) → f(x, g(y, z))
g(h(x, y), z) → g(x, f(y, z))
g(x, h(y, z)) → h(g(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof

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

G(h(x, y), z) → G(x, f(y, z))
G(f(x, y), z) → G(y, z)
G(x, h(y, z)) → G(x, y)

The TRS R consists of the following rules:

g(f(x, y), z) → f(x, g(y, z))
g(h(x, y), z) → g(x, f(y, z))
g(x, h(y, z)) → h(g(x, y), z)

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.


G(h(x, y), z) → G(x, f(y, z))
G(f(x, y), z) → G(y, z)
The remaining pairs can at least be oriented weakly.

G(x, h(y, z)) → G(x, y)
Used ordering: Combined order from the following AFS and order.
G(x1, x2)  =  G(x1)
h(x1, x2)  =  h(x1, x2)
f(x1, x2)  =  f(x1, x2)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof

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

G(x, h(y, z)) → G(x, y)

The TRS R consists of the following rules:

g(f(x, y), z) → f(x, g(y, z))
g(h(x, y), z) → g(x, f(y, z))
g(x, h(y, z)) → h(g(x, y), z)

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.


G(x, h(y, z)) → G(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
G(x1, x2)  =  G(x2)
h(x1, x2)  =  h(x1, x2)

Recursive Path Order [2].
Precedence:
h2 > G1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

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

g(f(x, y), z) → f(x, g(y, z))
g(h(x, y), z) → g(x, f(y, z))
g(x, h(y, z)) → h(g(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.