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:

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(times, 0), y) -> 0
app2(app2(times, app2(s, x)), y) -> app2(app2(plus, app2(app2(times, x), y)), y)
app2(inc, xs) -> app2(app2(map, app2(plus, app2(s, 0))), xs)
app2(double, xs) -> app2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(times, 0), y) -> 0
app2(app2(times, app2(s, x)), y) -> app2(app2(plus, app2(app2(times, x), y)), y)
app2(inc, xs) -> app2(app2(map, app2(plus, app2(s, 0))), xs)
app2(double, xs) -> app2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))

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:

APP2(app2(times, app2(s, x)), y) -> APP2(times, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(inc, xs) -> APP2(app2(map, app2(plus, app2(s, 0))), xs)
APP2(inc, xs) -> APP2(s, 0)
APP2(double, xs) -> APP2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(times, app2(s, x)), y) -> APP2(app2(plus, app2(app2(times, x), y)), y)
APP2(inc, xs) -> APP2(plus, app2(s, 0))
APP2(double, xs) -> APP2(map, app2(times, app2(s, app2(s, 0))))
APP2(double, xs) -> APP2(s, 0)
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(double, xs) -> APP2(s, app2(s, 0))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
APP2(double, xs) -> APP2(times, app2(s, app2(s, 0)))
APP2(app2(times, app2(s, x)), y) -> APP2(app2(times, x), y)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))
APP2(app2(times, app2(s, x)), y) -> APP2(plus, app2(app2(times, x), y))
APP2(inc, xs) -> APP2(map, app2(plus, app2(s, 0)))

The TRS R consists of the following rules:

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(times, 0), y) -> 0
app2(app2(times, app2(s, x)), y) -> app2(app2(plus, app2(app2(times, x), y)), y)
app2(inc, xs) -> app2(app2(map, app2(plus, app2(s, 0))), xs)
app2(double, xs) -> app2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))

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:

APP2(app2(times, app2(s, x)), y) -> APP2(times, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(s, app2(app2(plus, x), y))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(inc, xs) -> APP2(app2(map, app2(plus, app2(s, 0))), xs)
APP2(inc, xs) -> APP2(s, 0)
APP2(double, xs) -> APP2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(times, app2(s, x)), y) -> APP2(app2(plus, app2(app2(times, x), y)), y)
APP2(inc, xs) -> APP2(plus, app2(s, 0))
APP2(double, xs) -> APP2(map, app2(times, app2(s, app2(s, 0))))
APP2(double, xs) -> APP2(s, 0)
APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
APP2(double, xs) -> APP2(s, app2(s, 0))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(plus, app2(s, x)), y) -> APP2(plus, x)
APP2(double, xs) -> APP2(times, app2(s, app2(s, 0)))
APP2(app2(times, app2(s, x)), y) -> APP2(app2(times, x), y)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))
APP2(app2(times, app2(s, x)), y) -> APP2(plus, app2(app2(times, x), y))
APP2(inc, xs) -> APP2(map, app2(plus, app2(s, 0)))

The TRS R consists of the following rules:

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(times, 0), y) -> 0
app2(app2(times, app2(s, x)), y) -> app2(app2(plus, app2(app2(times, x), y)), y)
app2(inc, xs) -> app2(app2(map, app2(plus, app2(s, 0))), xs)
app2(double, xs) -> app2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 14 less nodes.

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

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

APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)

The TRS R consists of the following rules:

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(times, 0), y) -> 0
app2(app2(times, app2(s, x)), y) -> app2(app2(plus, app2(app2(times, x), y)), y)
app2(inc, xs) -> app2(app2(map, app2(plus, app2(s, 0))), xs)
app2(double, xs) -> app2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))

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.


APP2(app2(plus, app2(s, x)), y) -> APP2(app2(plus, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = max{0, x1 - 3}


POL( app2(x1, x2) ) = x2 + 2



The following usable rules [14] were oriented: none



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

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

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(times, 0), y) -> 0
app2(app2(times, app2(s, x)), y) -> app2(app2(plus, app2(app2(times, x), y)), y)
app2(inc, xs) -> app2(app2(map, app2(plus, app2(s, 0))), xs)
app2(double, xs) -> app2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))

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.

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

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

APP2(app2(times, app2(s, x)), y) -> APP2(app2(times, x), y)

The TRS R consists of the following rules:

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(times, 0), y) -> 0
app2(app2(times, app2(s, x)), y) -> app2(app2(plus, app2(app2(times, x), y)), y)
app2(inc, xs) -> app2(app2(map, app2(plus, app2(s, 0))), xs)
app2(double, xs) -> app2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))

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.


APP2(app2(times, app2(s, x)), y) -> APP2(app2(times, x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = max{0, x1 - 3}


POL( app2(x1, x2) ) = x2 + 2



The following usable rules [14] were oriented: none



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

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

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(times, 0), y) -> 0
app2(app2(times, app2(s, x)), y) -> app2(app2(plus, app2(app2(times, x), y)), y)
app2(inc, xs) -> app2(app2(map, app2(plus, app2(s, 0))), xs)
app2(double, xs) -> app2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))

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.

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

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

APP2(inc, xs) -> APP2(app2(map, app2(plus, app2(s, 0))), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(double, xs) -> APP2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)

The TRS R consists of the following rules:

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(times, 0), y) -> 0
app2(app2(times, app2(s, x)), y) -> app2(app2(plus, app2(app2(times, x), y)), y)
app2(inc, xs) -> app2(app2(map, app2(plus, app2(s, 0))), xs)
app2(double, xs) -> app2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))

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.


APP2(inc, xs) -> APP2(app2(map, app2(plus, app2(s, 0))), xs)
The remaining pairs can at least be oriented weakly.

APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(double, xs) -> APP2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = max{0, x1 - 2}


POL( inc ) = 3


POL( app2(x1, x2) ) = x2


POL( 0 ) = 0


POL( double ) = 0



The following usable rules [14] were oriented: none



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

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

APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(double, xs) -> APP2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)

The TRS R consists of the following rules:

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(times, 0), y) -> 0
app2(app2(times, app2(s, x)), y) -> app2(app2(plus, app2(app2(times, x), y)), y)
app2(inc, xs) -> app2(app2(map, app2(plus, app2(s, 0))), xs)
app2(double, xs) -> app2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))

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.


APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
The remaining pairs can at least be oriented weakly.

APP2(double, xs) -> APP2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = max{0, x2 - 3}


POL( app2(x1, x2) ) = x1 + x2 + 1


POL( cons ) = 2



The following usable rules [14] were oriented: none



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

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

APP2(double, xs) -> APP2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)

The TRS R consists of the following rules:

app2(app2(plus, 0), y) -> y
app2(app2(plus, app2(s, x)), y) -> app2(s, app2(app2(plus, x), y))
app2(app2(times, 0), y) -> 0
app2(app2(times, app2(s, x)), y) -> app2(app2(plus, app2(app2(times, x), y)), y)
app2(inc, xs) -> app2(app2(map, app2(plus, app2(s, 0))), xs)
app2(double, xs) -> app2(app2(map, app2(times, app2(s, app2(s, 0)))), xs)
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, x), xs)) -> app2(app2(cons, app2(f, x)), app2(app2(map, f), xs))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.