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(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)

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:

F1(f1(x)) -> F1(g2(f1(x), x))
F1(f1(x)) -> H2(f1(x), f1(x))
H2(x, x) -> G2(x, 0)
F1(f1(x)) -> G2(f1(x), x)
F1(f1(x)) -> F1(h2(f1(x), f1(x)))

The TRS R consists of the following rules:

f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)

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(f1(x)) -> F1(g2(f1(x), x))
F1(f1(x)) -> H2(f1(x), f1(x))
H2(x, x) -> G2(x, 0)
F1(f1(x)) -> G2(f1(x), x)
F1(f1(x)) -> F1(h2(f1(x), f1(x)))

The TRS R consists of the following rules:

f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)

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 3 less nodes.

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

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

F1(f1(x)) -> F1(g2(f1(x), x))
F1(f1(x)) -> F1(h2(f1(x), f1(x)))

The TRS R consists of the following rules:

f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)

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.


F1(f1(x)) -> F1(g2(f1(x), x))
The remaining pairs can at least be oriented weakly.

F1(f1(x)) -> F1(h2(f1(x), f1(x)))
Used ordering: Polynomial Order [17,21] with Interpretation:

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


POL( g2(x1, x2) ) = 2x2 + 2


POL( F1(x1) ) = 2x1 + 2


POL( h2(x1, x2) ) = 3


POL( f1(x1) ) = 2x1 + 3



The following usable rules [14] were oriented:

h2(x, x) -> g2(x, 0)
g2(x, y) -> y



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

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

F1(f1(x)) -> F1(h2(f1(x), f1(x)))

The TRS R consists of the following rules:

f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)

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.


F1(f1(x)) -> F1(h2(f1(x), f1(x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 0 ) = max{0, -2}


POL( g2(x1, x2) ) = x2 + 1


POL( F1(x1) ) = 2x1


POL( h2(x1, x2) ) = 1


POL( f1(x1) ) = 3x1 + 2



The following usable rules [14] were oriented:

h2(x, x) -> g2(x, 0)
g2(x, y) -> y



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

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

f1(f1(x)) -> f1(g2(f1(x), x))
f1(f1(x)) -> f1(h2(f1(x), f1(x)))
g2(x, y) -> y
h2(x, x) -> g2(x, 0)

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.