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(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
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))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, f), xs)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
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))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, 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(app2(app2(filter2, true), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(filter2, app2(f, x)), f)
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(filter, f)
APP2(app2(ack, app2(succ, x)), y) -> APP2(succ, 0)
APP2(app2(ack, 0), y) -> APP2(succ, y)
APP2(app2(ack, app2(succ, x)), y) -> APP2(app2(ack, x), app2(succ, 0))
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(filter2, app2(f, x)), f), x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(app2(cons, x), app2(app2(filter, f), xs))
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(filter, f)
APP2(app2(ack, app2(succ, x)), y) -> APP2(ack, x)
APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, app2(succ, x)), y)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(filter2, app2(f, x))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(cons, x)
APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(ack, x)
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))

The TRS R consists of the following rules:

app2(app2(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
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))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, 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(app2(app2(filter2, true), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(cons, app2(f, x)), app2(app2(map, f), xs))
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(filter2, app2(f, x)), f)
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(filter, f)
APP2(app2(ack, app2(succ, x)), y) -> APP2(succ, 0)
APP2(app2(ack, 0), y) -> APP2(succ, y)
APP2(app2(ack, app2(succ, x)), y) -> APP2(app2(ack, x), app2(succ, 0))
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(filter2, app2(f, x)), f), x)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(app2(cons, x), app2(app2(filter, f), xs))
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(filter, f)
APP2(app2(ack, app2(succ, x)), y) -> APP2(ack, x)
APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, app2(succ, x)), y)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(filter2, app2(f, x))
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(cons, x)
APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(ack, x)
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(cons, app2(f, x))

The TRS R consists of the following rules:

app2(app2(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
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))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, 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 2 SCCs with 14 less nodes.

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

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

APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, app2(succ, x)), y)
APP2(app2(ack, app2(succ, x)), y) -> APP2(app2(ack, x), app2(succ, 0))

The TRS R consists of the following rules:

app2(app2(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
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))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, 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(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
APP2(app2(ack, app2(succ, x)), y) -> APP2(app2(ack, x), app2(succ, 0))
The remaining pairs can at least be oriented weakly.

APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, app2(succ, x)), y)
Used ordering: Polynomial interpretation [21]:

POL(0) = 1   
POL(APP2(x1, x2)) = x1   
POL(ack) = 0   
POL(app2(x1, x2)) = 1 + x2   
POL(succ) = 1   

The following usable rules [14] were oriented: none



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

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

APP2(app2(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, app2(succ, x)), y)

The TRS R consists of the following rules:

app2(app2(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
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))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, 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(ack, app2(succ, x)), app2(succ, y)) -> APP2(app2(ack, app2(succ, x)), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = 2·x2   
POL(ack) = 0   
POL(app2(x1, x2)) = 1 + 2·x2   
POL(succ) = 2   

The following usable rules [14] were oriented: none



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

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

app2(app2(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
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))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, 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

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

APP2(app2(app2(app2(filter2, true), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(filter, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(app2(map, f), xs)

The TRS R consists of the following rules:

app2(app2(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
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))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, 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(app2(app2(filter2, true), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(map, f), app2(app2(cons, x), xs)) -> APP2(f, x)
APP2(app2(app2(app2(filter2, false), f), x), xs) -> APP2(app2(filter, f), xs)
APP2(app2(filter, 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.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = 2·x1 + 2·x2   
POL(app2(x1, x2)) = 2 + 2·x1 + x2   
POL(cons) = 0   
POL(false) = 1   
POL(filter) = 1   
POL(filter2) = 2   
POL(map) = 2   
POL(true) = 1   

The following usable rules [14] were oriented: none



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

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

app2(app2(ack, 0), y) -> app2(succ, y)
app2(app2(ack, app2(succ, x)), y) -> app2(app2(ack, x), app2(succ, 0))
app2(app2(ack, app2(succ, x)), app2(succ, y)) -> app2(app2(ack, x), app2(app2(ack, app2(succ, x)), y))
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))
app2(app2(filter, f), nil) -> nil
app2(app2(filter, f), app2(app2(cons, x), xs)) -> app2(app2(app2(app2(filter2, app2(f, x)), f), x), xs)
app2(app2(app2(app2(filter2, true), f), x), xs) -> app2(app2(cons, x), app2(app2(filter, f), xs))
app2(app2(app2(app2(filter2, false), f), x), xs) -> app2(app2(filter, 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.