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:

f(x, f(a, y)) → f(a, f(f(f(a, x), h(a)), y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(x, f(a, y)) → f(a, f(f(f(a, x), h(a)), y))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(x, f(a, y)) → F(a, x)
F(x, f(a, y)) → F(f(f(a, x), h(a)), y)
F(x, f(a, y)) → F(a, f(f(f(a, x), h(a)), y))
F(x, f(a, y)) → F(f(a, x), h(a))

The TRS R consists of the following rules:

f(x, f(a, y)) → f(a, f(f(f(a, x), h(a)), 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:

F(x, f(a, y)) → F(a, x)
F(x, f(a, y)) → F(f(f(a, x), h(a)), y)
F(x, f(a, y)) → F(a, f(f(f(a, x), h(a)), y))
F(x, f(a, y)) → F(f(a, x), h(a))

The TRS R consists of the following rules:

f(x, f(a, y)) → f(a, f(f(f(a, x), h(a)), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

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

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

F(x, f(a, y)) → F(a, x)
F(x, f(a, y)) → F(f(f(a, x), h(a)), y)
F(x, f(a, y)) → F(a, f(f(f(a, x), h(a)), y))

The TRS R consists of the following rules:

f(x, f(a, y)) → f(a, f(f(f(a, x), h(a)), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(x, f(a, y)) → F(a, x)
The remaining pairs can at least be oriented weakly.

F(x, f(a, y)) → F(f(f(a, x), h(a)), y)
F(x, f(a, y)) → F(a, f(f(f(a, x), h(a)), y))
Used ordering: Polynomial interpretation [25,35]:

POL(a) = 0   
POL(f(x1, x2)) = 1 + x_2   
POL(h(x1)) = 0   
POL(F(x1, x2)) = (4)x_1 + (4)x_2   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

f(x, f(a, y)) → f(a, f(f(f(a, x), h(a)), y))



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

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

F(x, f(a, y)) → F(f(f(a, x), h(a)), y)
F(x, f(a, y)) → F(a, f(f(f(a, x), h(a)), y))

The TRS R consists of the following rules:

f(x, f(a, y)) → f(a, f(f(f(a, x), h(a)), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(x, f(a, y)) → F(f(f(a, x), h(a)), y)
F(x, f(a, y)) → F(a, f(f(f(a, x), h(a)), y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(a) = 4   
POL(f(x1, x2)) = (1/4)x_1 + (2)x_2   
POL(h(x1)) = 0   
POL(F(x1, x2)) = (1/4)x_1 + (2)x_2   
The value of delta used in the strict ordering is 7/8.
The following usable rules [17] were oriented:

f(x, f(a, y)) → f(a, f(f(f(a, x), h(a)), 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:

f(x, f(a, y)) → f(a, f(f(f(a, x), h(a)), 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.