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(append, nil), l) -> l
app2(app2(append, app2(app2(cons, h), t)), l) -> app2(app2(cons, h), app2(app2(append, t), l))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(append, app2(app2(append, l1), l2)), l3) -> app2(app2(append, l1), app2(app2(append, l2), l3))
app2(app2(map, f), app2(app2(append, l1), l2)) -> app2(app2(append, app2(app2(map, f), l1)), app2(app2(map, f), l2))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app2(app2(append, nil), l) -> l
app2(app2(append, app2(app2(cons, h), t)), l) -> app2(app2(cons, h), app2(app2(append, t), l))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(append, app2(app2(append, l1), l2)), l3) -> app2(app2(append, l1), app2(app2(append, l2), l3))
app2(app2(map, f), app2(app2(append, l1), l2)) -> app2(app2(append, app2(app2(map, f), l1)), app2(app2(map, f), l2))

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(append, app2(app2(append, l1), l2)), l3) -> APP2(app2(append, l2), l3)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(app2(cons, app2(f, h)), app2(app2(map, f), t))
APP2(app2(append, app2(app2(cons, h), t)), l) -> APP2(append, t)
APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(app2(append, app2(app2(map, f), l1)), app2(app2(map, f), l2))
APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(append, app2(app2(map, f), l1))
APP2(app2(append, app2(app2(append, l1), l2)), l3) -> APP2(app2(append, l1), app2(app2(append, l2), l3))
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(app2(map, f), t)
APP2(app2(append, app2(app2(append, l1), l2)), l3) -> APP2(append, l2)
APP2(app2(append, app2(app2(cons, h), t)), l) -> APP2(app2(append, t), l)
APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(app2(map, f), l1)
APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(app2(map, f), l2)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(f, h)
APP2(app2(append, app2(app2(cons, h), t)), l) -> APP2(app2(cons, h), app2(app2(append, t), l))
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(cons, app2(f, h))

The TRS R consists of the following rules:

app2(app2(append, nil), l) -> l
app2(app2(append, app2(app2(cons, h), t)), l) -> app2(app2(cons, h), app2(app2(append, t), l))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(append, app2(app2(append, l1), l2)), l3) -> app2(app2(append, l1), app2(app2(append, l2), l3))
app2(app2(map, f), app2(app2(append, l1), l2)) -> app2(app2(append, app2(app2(map, f), l1)), app2(app2(map, f), l2))

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(append, app2(app2(append, l1), l2)), l3) -> APP2(app2(append, l2), l3)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(app2(cons, app2(f, h)), app2(app2(map, f), t))
APP2(app2(append, app2(app2(cons, h), t)), l) -> APP2(append, t)
APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(app2(append, app2(app2(map, f), l1)), app2(app2(map, f), l2))
APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(append, app2(app2(map, f), l1))
APP2(app2(append, app2(app2(append, l1), l2)), l3) -> APP2(app2(append, l1), app2(app2(append, l2), l3))
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(app2(map, f), t)
APP2(app2(append, app2(app2(append, l1), l2)), l3) -> APP2(append, l2)
APP2(app2(append, app2(app2(cons, h), t)), l) -> APP2(app2(append, t), l)
APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(app2(map, f), l1)
APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(app2(map, f), l2)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(f, h)
APP2(app2(append, app2(app2(cons, h), t)), l) -> APP2(app2(cons, h), app2(app2(append, t), l))
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(cons, app2(f, h))

The TRS R consists of the following rules:

app2(app2(append, nil), l) -> l
app2(app2(append, app2(app2(cons, h), t)), l) -> app2(app2(cons, h), app2(app2(append, t), l))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(append, app2(app2(append, l1), l2)), l3) -> app2(app2(append, l1), app2(app2(append, l2), l3))
app2(app2(map, f), app2(app2(append, l1), l2)) -> app2(app2(append, app2(app2(map, f), l1)), app2(app2(map, f), l2))

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 7 less nodes.

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

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

APP2(app2(append, app2(app2(append, l1), l2)), l3) -> APP2(app2(append, l2), l3)
APP2(app2(append, app2(app2(append, l1), l2)), l3) -> APP2(app2(append, l1), app2(app2(append, l2), l3))
APP2(app2(append, app2(app2(cons, h), t)), l) -> APP2(app2(append, t), l)

The TRS R consists of the following rules:

app2(app2(append, nil), l) -> l
app2(app2(append, app2(app2(cons, h), t)), l) -> app2(app2(cons, h), app2(app2(append, t), l))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(append, app2(app2(append, l1), l2)), l3) -> app2(app2(append, l1), app2(app2(append, l2), l3))
app2(app2(map, f), app2(app2(append, l1), l2)) -> app2(app2(append, app2(app2(map, f), l1)), app2(app2(map, f), l2))

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 strictly oriented and are deleted.


APP2(app2(append, app2(app2(append, l1), l2)), l3) -> APP2(app2(append, l2), l3)
APP2(app2(append, app2(app2(append, l1), l2)), l3) -> APP2(app2(append, l1), app2(app2(append, l2), l3))
APP2(app2(append, app2(app2(cons, h), t)), l) -> APP2(app2(append, t), l)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP2(x1, x2)
app2(x1, x2)  =  app2(x1, x2)
append  =  append
cons  =  cons
nil  =  nil

Lexicographic Path Order [19].
Precedence:
cons > [APP2, append] > app2
nil > app2


The following usable rules [14] were oriented:

app2(app2(append, nil), l) -> l
app2(app2(append, app2(app2(cons, h), t)), l) -> app2(app2(cons, h), app2(app2(append, t), l))
app2(app2(append, app2(app2(append, l1), l2)), l3) -> app2(app2(append, l1), app2(app2(append, l2), l3))



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

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

app2(app2(append, nil), l) -> l
app2(app2(append, app2(app2(cons, h), t)), l) -> app2(app2(cons, h), app2(app2(append, t), l))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(append, app2(app2(append, l1), l2)), l3) -> app2(app2(append, l1), app2(app2(append, l2), l3))
app2(app2(map, f), app2(app2(append, l1), l2)) -> app2(app2(append, app2(app2(map, f), l1)), app2(app2(map, f), l2))

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(map, f), app2(app2(append, l1), l2)) -> APP2(app2(map, f), l1)
APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(app2(map, f), l2)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(f, h)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(app2(map, f), t)

The TRS R consists of the following rules:

app2(app2(append, nil), l) -> l
app2(app2(append, app2(app2(cons, h), t)), l) -> app2(app2(cons, h), app2(app2(append, t), l))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(append, app2(app2(append, l1), l2)), l3) -> app2(app2(append, l1), app2(app2(append, l2), l3))
app2(app2(map, f), app2(app2(append, l1), l2)) -> app2(app2(append, app2(app2(map, f), l1)), app2(app2(map, f), l2))

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 strictly oriented and are deleted.


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

APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(app2(map, f), l1)
APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(app2(map, f), l2)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(app2(map, f), t)
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  x1
app2(x1, x2)  =  app1(x2)
map  =  map
append  =  append
cons  =  cons

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(app2(map, f), l1)
APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(app2(map, f), l2)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(app2(map, f), t)

The TRS R consists of the following rules:

app2(app2(append, nil), l) -> l
app2(app2(append, app2(app2(cons, h), t)), l) -> app2(app2(cons, h), app2(app2(append, t), l))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(append, app2(app2(append, l1), l2)), l3) -> app2(app2(append, l1), app2(app2(append, l2), l3))
app2(app2(map, f), app2(app2(append, l1), l2)) -> app2(app2(append, app2(app2(map, f), l1)), app2(app2(map, f), l2))

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 strictly oriented and are deleted.


APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(app2(map, f), l1)
APP2(app2(map, f), app2(app2(append, l1), l2)) -> APP2(app2(map, f), l2)
APP2(app2(map, f), app2(app2(cons, h), t)) -> APP2(app2(map, f), t)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x2)
app2(x1, x2)  =  app2(x1, x2)
map  =  map
append  =  append
cons  =  cons

Lexicographic Path Order [19].
Precedence:
append > app2 > APP1
cons > app2 > APP1


The following usable rules [14] were oriented: none



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

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

app2(app2(append, nil), l) -> l
app2(app2(append, app2(app2(cons, h), t)), l) -> app2(app2(cons, h), app2(app2(append, t), l))
app2(app2(map, f), nil) -> nil
app2(app2(map, f), app2(app2(cons, h), t)) -> app2(app2(cons, app2(f, h)), app2(app2(map, f), t))
app2(app2(append, app2(app2(append, l1), l2)), l3) -> app2(app2(append, l1), app2(app2(append, l2), l3))
app2(app2(map, f), app2(app2(append, l1), l2)) -> app2(app2(append, app2(app2(map, f), l1)), app2(app2(map, f), l2))

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.