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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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
      ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


G2(x, h2(y, z)) -> G2(x, y)
The remaining pairs can at least be oriented weakly.

G2(h2(x, y), z) -> G2(x, f2(y, z))
G2(f2(x, y), z) -> G2(y, z)
Used ordering: Polynomial interpretation [21]:

POL(G2(x1, x2)) = x2   
POL(f2(x1, x2)) = 0   
POL(h2(x1, x2)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


G2(h2(x, y), z) -> G2(x, f2(y, z))
The remaining pairs can at least be oriented weakly.

G2(f2(x, y), z) -> G2(y, z)
Used ordering: Polynomial interpretation [21]:

POL(G2(x1, x2)) = x1   
POL(f2(x1, x2)) = x2   
POL(h2(x1, x2)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


G2(f2(x, y), z) -> G2(y, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(G2(x1, x2)) = x1   
POL(f2(x1, x2)) = 1 + x2   

The following usable rules [14] 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:

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.