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:

app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, x)), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, x)), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(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:

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, x), app(app(ack, app(succ, x)), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(ack, app(succ, x)), y) → APP(succ, 0)
APP(app(ack, 0), y) → APP(succ, y)
APP(app(ack, app(succ, x)), y) → APP(app(ack, x), app(succ, 0))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(ack, app(succ, x)), y) → APP(ack, x)
APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, app(succ, x)), y)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(ack, app(succ, x)), app(succ, y)) → APP(ack, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, x)), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, x), app(app(ack, app(succ, x)), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(ack, app(succ, x)), y) → APP(succ, 0)
APP(app(ack, 0), y) → APP(succ, y)
APP(app(ack, app(succ, x)), y) → APP(app(ack, x), app(succ, 0))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(ack, app(succ, x)), y) → APP(ack, x)
APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, app(succ, x)), y)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(ack, app(succ, x)), app(succ, y)) → APP(ack, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, x)), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, x), app(app(ack, app(succ, x)), y))
APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(filter2, app(f, x)), f)
APP(app(ack, app(succ, x)), y) → APP(succ, 0)
APP(app(app(app(filter2, false), f), x), xs) → APP(filter, f)
APP(app(ack, app(succ, x)), y) → APP(app(ack, x), app(succ, 0))
APP(app(ack, 0), y) → APP(succ, y)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(filter2, app(f, x)), f), x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(app(app(app(filter2, app(f, x)), f), x), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(app(cons, x), app(app(filter, f), xs))
APP(app(app(app(filter2, true), f), x), xs) → APP(filter, f)
APP(app(ack, app(succ, x)), y) → APP(ack, x)
APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, app(succ, x)), y)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(filter2, app(f, x))
APP(app(ack, app(succ, x)), app(succ, y)) → APP(ack, x)
APP(app(app(app(filter2, true), f), x), xs) → APP(cons, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))

The TRS R consists of the following rules:

app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, x)), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, x), app(app(ack, app(succ, x)), y))
APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, app(succ, x)), y)
APP(app(ack, app(succ, x)), y) → APP(app(ack, x), app(succ, 0))

The TRS R consists of the following rules:

app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, x)), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

ACK(succ(x), y) → ACK(x, succ(0))
ACK(succ(x), succ(y)) → ACK(x, ack(succ(x), y))
ACK(succ(x), succ(y)) → ACK(succ(x), y)

The TRS R consists of the following rules:

ack(succ(x), succ(y)) → ack(x, ack(succ(x), y))
ack(succ(x), y) → ack(x, succ(0))
ack(0, y) → succ(y)

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


The following pairs can be oriented strictly and are deleted.


APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, x), app(app(ack, app(succ, x)), y))
APP(app(ack, app(succ, x)), y) → APP(app(ack, x), app(succ, 0))
The remaining pairs can at least be oriented weakly.

APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, app(succ, x)), y)
Used ordering: Combined order from the following AFS and order.
ACK(x1, x2)  =  x1
succ(x1)  =  succ(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, app(succ, x)), y)

The TRS R consists of the following rules:

app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, x)), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(filter, f), xs)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13]. Here, we combined the reduction pair processor with the A-transformation [14] which results in the following intermediate Q-DP Problem.
Q DP problem:
The TRS P consists of the following rules:

ACK(succ(x), succ(y)) → ACK(succ(x), y)

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


The following pairs can be oriented strictly and are deleted.


APP(app(ack, app(succ, x)), app(succ, y)) → APP(app(ack, app(succ, x)), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
ACK(x1, x2)  =  x2
succ(x1)  =  succ(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, x)), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)

The TRS R consists of the following rules:

app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, x)), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(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.


APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(filter, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
The remaining pairs can at least be oriented weakly.

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x2
app(x1, x2)  =  app(x1, x2)
cons  =  cons

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

APP(app(app(app(filter2, true), f), x), xs) → APP(app(filter, f), xs)
APP(app(app(app(filter2, false), f), x), xs) → APP(app(filter, f), xs)

The TRS R consists of the following rules:

app(app(ack, 0), y) → app(succ, y)
app(app(ack, app(succ, x)), y) → app(app(ack, x), app(succ, 0))
app(app(ack, app(succ, x)), app(succ, y)) → app(app(ack, x), app(app(ack, app(succ, x)), y))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, x), xs)) → app(app(cons, app(f, x)), app(app(map, f), xs))
app(app(filter, f), nil) → nil
app(app(filter, f), app(app(cons, x), xs)) → app(app(app(app(filter2, app(f, x)), f), x), xs)
app(app(app(app(filter2, true), f), x), xs) → app(app(cons, x), app(app(filter, f), xs))
app(app(app(app(filter2, false), f), x), xs) → app(app(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 0 SCCs with 2 less nodes.