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:

f2(a, g1(y)) -> g1(g1(y))
f2(g1(x), a) -> f2(x, g1(a))
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
h3(a, y, z) -> z

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f2(a, g1(y)) -> g1(g1(y))
f2(g1(x), a) -> f2(x, g1(a))
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
h3(a, y, z) -> 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:

H3(g1(x), y, z) -> H3(x, y, z)
H3(g1(x), y, z) -> F2(y, h3(x, y, z))
F2(g1(x), g1(y)) -> H3(g1(y), x, g1(y))
F2(g1(x), a) -> F2(x, g1(a))

The TRS R consists of the following rules:

f2(a, g1(y)) -> g1(g1(y))
f2(g1(x), a) -> f2(x, g1(a))
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
h3(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:

H3(g1(x), y, z) -> H3(x, y, z)
H3(g1(x), y, z) -> F2(y, h3(x, y, z))
F2(g1(x), g1(y)) -> H3(g1(y), x, g1(y))
F2(g1(x), a) -> F2(x, g1(a))

The TRS R consists of the following rules:

f2(a, g1(y)) -> g1(g1(y))
f2(g1(x), a) -> f2(x, g1(a))
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
h3(a, y, z) -> 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.


F2(g1(x), a) -> F2(x, g1(a))
The remaining pairs can at least be oriented weakly.

H3(g1(x), y, z) -> H3(x, y, z)
H3(g1(x), y, z) -> F2(y, h3(x, y, z))
F2(g1(x), g1(y)) -> H3(g1(y), x, g1(y))
Used ordering: Polynomial interpretation [21]:

POL(F2(x1, x2)) = x2   
POL(H3(x1, x2, x3)) = x3   
POL(a) = 1   
POL(f2(x1, x2)) = 0   
POL(g1(x1)) = 0   
POL(h3(x1, x2, x3)) = x3   

The following usable rules [14] were oriented:

h3(a, y, z) -> z
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
f2(g1(x), a) -> f2(x, g1(a))
f2(a, g1(y)) -> g1(g1(y))



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

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

F2(g1(x), g1(y)) -> H3(g1(y), x, g1(y))
H3(g1(x), y, z) -> F2(y, h3(x, y, z))
H3(g1(x), y, z) -> H3(x, y, z)

The TRS R consists of the following rules:

f2(a, g1(y)) -> g1(g1(y))
f2(g1(x), a) -> f2(x, g1(a))
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
h3(a, y, z) -> 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.


F2(g1(x), g1(y)) -> H3(g1(y), x, g1(y))
The remaining pairs can at least be oriented weakly.

H3(g1(x), y, z) -> F2(y, h3(x, y, z))
H3(g1(x), y, z) -> H3(x, y, z)
Used ordering: Polynomial interpretation [21]:

POL(F2(x1, x2)) = x1   
POL(H3(x1, x2, x3)) = x2   
POL(a) = 0   
POL(f2(x1, x2)) = 0   
POL(g1(x1)) = 1 + x1   
POL(h3(x1, x2, x3)) = 0   

The following usable rules [14] were oriented: none



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

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

H3(g1(x), y, z) -> H3(x, y, z)
H3(g1(x), y, z) -> F2(y, h3(x, y, z))

The TRS R consists of the following rules:

f2(a, g1(y)) -> g1(g1(y))
f2(g1(x), a) -> f2(x, g1(a))
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
h3(a, y, z) -> z

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

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

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

H3(g1(x), y, z) -> H3(x, y, z)

The TRS R consists of the following rules:

f2(a, g1(y)) -> g1(g1(y))
f2(g1(x), a) -> f2(x, g1(a))
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
h3(a, y, z) -> 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.


H3(g1(x), y, z) -> H3(x, y, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(H3(x1, x2, x3)) = x1   
POL(g1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ PisEmptyProof

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

f2(a, g1(y)) -> g1(g1(y))
f2(g1(x), a) -> f2(x, g1(a))
f2(g1(x), g1(y)) -> h3(g1(y), x, g1(y))
h3(g1(x), y, z) -> f2(y, h3(x, y, z))
h3(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.