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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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

The set Q consists of the following terms:

f2(a, g1(x0))
f2(g1(x0), a)
f2(g1(x0), g1(x1))
h3(g1(x0), x1, x2)
h3(a, x0, x1)


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

The set Q consists of the following terms:

f2(a, g1(x0))
f2(g1(x0), a)
f2(g1(x0), g1(x1))
h3(g1(x0), x1, x2)
h3(a, x0, x1)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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

The set Q consists of the following terms:

f2(a, g1(x0))
f2(g1(x0), a)
f2(g1(x0), g1(x1))
h3(g1(x0), x1, x2)
h3(a, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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

H3(g1(x), y, z) -> H3(x, y, z)
H3(g1(x), y, z) -> F2(y, h3(x, y, z))
Used ordering: Combined order from the following AFS and order.
H3(x1, x2, x3)  =  H1(x2)
g1(x1)  =  g1(x1)
F2(x1, x2)  =  F1(x1)
h3(x1, x2, x3)  =  h2(x1, x3)
a  =  a
f2(x1, x2)  =  x1

Lexicographic Path Order [19].
Precedence:
g1 > h2 > [H1, F1]
g1 > a > [H1, F1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

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

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

The set Q consists of the following terms:

f2(a, g1(x0))
f2(g1(x0), a)
f2(g1(x0), g1(x1))
h3(g1(x0), x1, x2)
h3(a, x0, x1)

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
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ 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

The set Q consists of the following terms:

f2(a, g1(x0))
f2(g1(x0), a)
f2(g1(x0), g1(x1))
h3(g1(x0), x1, x2)
h3(a, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


H3(g1(x), y, z) -> H3(x, y, z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
H3(x1, x2, x3)  =  H1(x1)
g1(x1)  =  g1(x1)

Lexicographic Path Order [19].
Precedence:
[H1, g1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ 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

The set Q consists of the following terms:

f2(a, g1(x0))
f2(g1(x0), a)
f2(g1(x0), g1(x1))
h3(g1(x0), x1, x2)
h3(a, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.