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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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(x, f2(a, y)) -> F2(a, f2(f2(a, h1(f2(a, x))), y))
F2(x, f2(a, y)) -> F2(f2(a, h1(f2(a, x))), y)
F2(x, f2(a, y)) -> F2(a, h1(f2(a, x)))
F2(x, f2(a, y)) -> F2(a, x)

The TRS R consists of the following rules:

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

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(x, f2(a, y)) -> F2(a, f2(f2(a, h1(f2(a, x))), y))
F2(x, f2(a, y)) -> F2(f2(a, h1(f2(a, x))), y)
F2(x, f2(a, y)) -> F2(a, h1(f2(a, x)))
F2(x, f2(a, y)) -> F2(a, x)

The TRS R consists of the following rules:

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

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 1 less node.

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

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

F2(x, f2(a, y)) -> F2(a, f2(f2(a, h1(f2(a, x))), y))
F2(x, f2(a, y)) -> F2(f2(a, h1(f2(a, x))), y)
F2(x, f2(a, y)) -> F2(a, x)

The TRS R consists of the following rules:

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

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(x, f2(a, y)) -> F2(a, x)
The remaining pairs can at least be oriented weakly.

F2(x, f2(a, y)) -> F2(a, f2(f2(a, h1(f2(a, x))), y))
F2(x, f2(a, y)) -> F2(f2(a, h1(f2(a, x))), y)
Used ordering: Polynomial Order [17,21] with Interpretation:

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


POL( f2(x1, x2) ) = x2 + 3


POL( a ) = 0


POL( h1(x1) ) = 0



The following usable rules [14] were oriented:

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



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

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

F2(x, f2(a, y)) -> F2(a, f2(f2(a, h1(f2(a, x))), y))
F2(x, f2(a, y)) -> F2(f2(a, h1(f2(a, x))), y)

The TRS R consists of the following rules:

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

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(x, f2(a, y)) -> F2(a, f2(f2(a, h1(f2(a, x))), y))
F2(x, f2(a, y)) -> F2(f2(a, h1(f2(a, x))), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F2(x1, x2) ) = max{0, x2 - 1}


POL( f2(x1, x2) ) = max{0, x1 + 3x2 - 1}


POL( a ) = 3


POL( h1(x1) ) = 0



The following usable rules [14] were oriented:

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

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

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.