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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

POL(F2(x1, x2)) = x1·x2 + 3·x2   
POL(a) = 1   
POL(f2(x1, x2)) = 3·x1·x2 + 3·x2   
POL(h1(x1)) = 0   

The following usable rules [14] were oriented:

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



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

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

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