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:

g1(h1(g1(x))) -> g1(x)
g1(g1(x)) -> g1(h1(g1(x)))
h1(h1(x)) -> h1(f2(h1(x), x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

g1(h1(g1(x))) -> g1(x)
g1(g1(x)) -> g1(h1(g1(x)))
h1(h1(x)) -> h1(f2(h1(x), x))

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:

G1(g1(x)) -> H1(g1(x))
G1(g1(x)) -> G1(h1(g1(x)))
H1(h1(x)) -> H1(f2(h1(x), x))

The TRS R consists of the following rules:

g1(h1(g1(x))) -> g1(x)
g1(g1(x)) -> g1(h1(g1(x)))
h1(h1(x)) -> h1(f2(h1(x), 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:

G1(g1(x)) -> H1(g1(x))
G1(g1(x)) -> G1(h1(g1(x)))
H1(h1(x)) -> H1(f2(h1(x), x))

The TRS R consists of the following rules:

g1(h1(g1(x))) -> g1(x)
g1(g1(x)) -> g1(h1(g1(x)))
h1(h1(x)) -> h1(f2(h1(x), x))

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 2 less nodes.

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

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

G1(g1(x)) -> G1(h1(g1(x)))

The TRS R consists of the following rules:

g1(h1(g1(x))) -> g1(x)
g1(g1(x)) -> g1(h1(g1(x)))
h1(h1(x)) -> h1(f2(h1(x), x))

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.


G1(g1(x)) -> G1(h1(g1(x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( h1(x1) ) = max{0, -1}


POL( G1(x1) ) = max{0, 2x1 - 3}


POL( g1(x1) ) = 3x1 + 2


POL( f2(x1, x2) ) = 1



The following usable rules [14] were oriented:

h1(h1(x)) -> h1(f2(h1(x), x))



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

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

g1(h1(g1(x))) -> g1(x)
g1(g1(x)) -> g1(h1(g1(x)))
h1(h1(x)) -> h1(f2(h1(x), 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.