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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

F2(a, x) -> G1(x)
F2(a, x) -> F2(g1(x), x)
G1(h1(x)) -> G1(x)
H1(g1(x)) -> H1(a)

The TRS R consists of the following rules:

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

F2(a, x) -> G1(x)
F2(a, x) -> F2(g1(x), x)
G1(h1(x)) -> G1(x)
H1(g1(x)) -> H1(a)

The TRS R consists of the following rules:

f2(a, x) -> f2(g1(x), x)
h1(g1(x)) -> h1(a)
g1(h1(x)) -> g1(x)
h1(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 2 SCCs with 2 less nodes.

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

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

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

The TRS R consists of the following rules:

f2(a, x) -> f2(g1(x), x)
h1(g1(x)) -> h1(a)
g1(h1(x)) -> g1(x)
h1(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(h1(x)) -> G1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( h1(x1) ) = 2x1 + 1


POL( G1(x1) ) = x1 + 2



The following usable rules [14] were oriented: none



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

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

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

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

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

F2(a, x) -> F2(g1(x), x)

The TRS R consists of the following rules:

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


F2(a, x) -> F2(g1(x), x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

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


POL( a ) = 2


POL( F2(x1, x2) ) = max{0, 2x1 + x2 - 2}


POL( g1(x1) ) = max{0, -1}



The following usable rules [14] were oriented:

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



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

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

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