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:

ap(ap(ff, x), x) → ap(ap(x, ap(ff, x)), ap(ap(cons, x), nil))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

ap(ap(ff, x), x) → ap(ap(x, ap(ff, x)), ap(ap(cons, x), nil))

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:

AP(ap(ff, x), x) → AP(x, ap(ff, x))
AP(ap(ff, x), x) → AP(cons, x)
AP(ap(ff, x), x) → AP(ap(x, ap(ff, x)), ap(ap(cons, x), nil))
AP(ap(ff, x), x) → AP(ap(cons, x), nil)

The TRS R consists of the following rules:

ap(ap(ff, x), x) → ap(ap(x, ap(ff, x)), ap(ap(cons, x), nil))

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:

AP(ap(ff, x), x) → AP(x, ap(ff, x))
AP(ap(ff, x), x) → AP(cons, x)
AP(ap(ff, x), x) → AP(ap(x, ap(ff, x)), ap(ap(cons, x), nil))
AP(ap(ff, x), x) → AP(ap(cons, x), nil)

The TRS R consists of the following rules:

ap(ap(ff, x), x) → ap(ap(x, ap(ff, x)), ap(ap(cons, x), nil))

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 2 less nodes.

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

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

AP(ap(ff, x), x) → AP(x, ap(ff, x))
AP(ap(ff, x), x) → AP(ap(x, ap(ff, x)), ap(ap(cons, x), nil))

The TRS R consists of the following rules:

ap(ap(ff, x), x) → ap(ap(x, ap(ff, x)), ap(ap(cons, x), nil))

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.


AP(ap(ff, x), x) → AP(ap(x, ap(ff, x)), ap(ap(cons, x), nil))
The remaining pairs can at least be oriented weakly.

AP(ap(ff, x), x) → AP(x, ap(ff, x))
Used ordering: Polynomial interpretation [25,35]:

POL(ap(x1, x2)) = (1/4)x_1 + (1/4)x_2   
POL(cons) = 0   
POL(AP(x1, x2)) = (1/4)x_1 + (1/4)x_2   
POL(ff) = 4   
POL(nil) = 0   
The value of delta used in the strict ordering is 3/16.
The following usable rules [17] were oriented:

ap(ap(ff, x), x) → ap(ap(x, ap(ff, x)), ap(ap(cons, x), nil))



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

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

AP(ap(ff, x), x) → AP(x, ap(ff, x))

The TRS R consists of the following rules:

ap(ap(ff, x), x) → ap(ap(x, ap(ff, x)), ap(ap(cons, x), nil))

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.


AP(ap(ff, x), x) → AP(x, ap(ff, x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(ap(x1, x2)) = 1/2 + (2)x_2   
POL(AP(x1, x2)) = (1/2)x_1   
POL(ff) = 0   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented: none



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

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

ap(ap(ff, x), x) → ap(ap(x, ap(ff, x)), ap(ap(cons, x), nil))

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.