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:

f1(a) -> g1(h1(a))
h1(g1(x)) -> g1(h1(f1(x)))
k3(x, h1(x), a) -> h1(x)
k3(f1(x), y, x) -> f1(x)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(a) -> g1(h1(a))
h1(g1(x)) -> g1(h1(f1(x)))
k3(x, h1(x), a) -> h1(x)
k3(f1(x), y, x) -> f1(x)

Q is empty.

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

F1(a) -> H1(a)
H1(g1(x)) -> F1(x)
H1(g1(x)) -> H1(f1(x))

The TRS R consists of the following rules:

f1(a) -> g1(h1(a))
h1(g1(x)) -> g1(h1(f1(x)))
k3(x, h1(x), a) -> h1(x)
k3(f1(x), y, x) -> f1(x)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

F1(a) -> H1(a)
H1(g1(x)) -> F1(x)
H1(g1(x)) -> H1(f1(x))

The TRS R consists of the following rules:

f1(a) -> g1(h1(a))
h1(g1(x)) -> g1(h1(f1(x)))
k3(x, h1(x), a) -> h1(x)
k3(f1(x), y, x) -> f1(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof

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

H1(g1(x)) -> H1(f1(x))

The TRS R consists of the following rules:

f1(a) -> g1(h1(a))
h1(g1(x)) -> g1(h1(f1(x)))
k3(x, h1(x), a) -> h1(x)
k3(f1(x), y, x) -> f1(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

H1(g1(x)) -> H1(f1(x))
Used argument filtering: H1(x1)  =  x1
g1(x1)  =  g1(x1)
f1(x1)  =  x1
a  =  a
h1(x1)  =  h
Used ordering: Precedence:
a > g1
a > h



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ PisEmptyProof

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

f1(a) -> g1(h1(a))
h1(g1(x)) -> g1(h1(f1(x)))
k3(x, h1(x), a) -> h1(x)
k3(f1(x), y, x) -> f1(x)

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.