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 [1,15] 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
      ↳ 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 [15].


The following pairs can be oriented strictly and are deleted.


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

G(f(x, y), z) → G(y, z)
G(x, h(y, z)) → G(x, y)
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1, x2)) = (3/4)x_1 + (2)x_2   
POL(h(x1, x2)) = 1/4 + (3/2)x_1   
POL(G(x1, x2)) = (3/2)x_1   
The value of delta used in the strict ordering is 3/8.
The following usable rules [17] were oriented: none



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

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

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 [15].


The following pairs can be oriented strictly and are deleted.


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: Polynomial interpretation [25,35]:

POL(f(x1, x2)) = 1 + (11/4)x_2   
POL(h(x1, x2)) = (4)x_1   
POL(G(x1, x2)) = (4)x_1 + (3/2)x_2   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ 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 [15].


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: Polynomial interpretation [25,35]:

POL(h(x1, x2)) = 1/4 + (7/2)x_1   
POL(G(x1, x2)) = (2)x_2   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ 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.