Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

The TRS R consists of the following rules:

f(x, f(a, y)) → f(a, f(f(a, h(f(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:

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

POL(a) = 0   
POL(f(x1, x2)) = 4 + (2)x_2   
POL(h(x1)) = 0   
POL(F(x1, x2)) = (3)x_1 + (3)x_2   
The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented:

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

The TRS R consists of the following rules:

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

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

POL(a) = 0   
POL(f(x1, x2)) = 4 + (4)x_2   
POL(h(x1)) = 0   
POL(F(x1, x2)) = (4)x_2   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented:

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



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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.