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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

H(g(x), y, z) → F(y, h(x, y, z))
H(g(x), y, z) → H(x, y, z)
F(g(x), a) → F(x, g(a))
F(g(x), g(y)) → H(g(y), x, g(y))

The TRS R consists of the following rules:

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

H(g(x), y, z) → F(y, h(x, y, z))
H(g(x), y, z) → H(x, y, z)
F(g(x), a) → F(x, g(a))
F(g(x), g(y)) → H(g(y), x, g(y))

The TRS R consists of the following rules:

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


H(g(x), y, z) → F(y, h(x, y, z))
F(g(x), a) → F(x, g(a))
F(g(x), g(y)) → H(g(y), x, g(y))
The remaining pairs can at least be oriented weakly.

H(g(x), y, z) → H(x, y, z)
Used ordering: Polynomial interpretation [25,35]:

POL(a) = 0   
POL(g(x1)) = 4 + x_1   
POL(h(x1, x2, x3)) = 2 + (2)x_1 + (2)x_2   
POL(f(x1, x2)) = 2 + (4)x_1   
POL(F(x1, x2)) = 1 + (2)x_1   
POL(H(x1, x2, x3)) = 2 + (2)x_2   
The value of delta used in the strict ordering is 1.
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:

H(g(x), y, z) → H(x, y, z)

The TRS R consists of the following rules:

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


H(g(x), y, z) → H(x, y, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(g(x1)) = 1 + (4)x_1   
POL(H(x1, x2, x3)) = (4)x_1   
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
              ↳ PisEmptyProof

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

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