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(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(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:

APP(app(append, app(app(append, l1), l2)), l3) → APP(app(append, l2), l3)
APP(app(map, f), app(app(cons, h), t)) → APP(app(cons, app(f, h)), app(app(map, f), t))
APP(app(append, app(app(cons, h), t)), l) → APP(append, t)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(append, app(app(map, f), l1)), app(app(map, f), l2))
APP(app(map, f), app(app(append, l1), l2)) → APP(append, app(app(map, f), l1))
APP(app(append, app(app(append, l1), l2)), l3) → APP(app(append, l1), app(app(append, l2), l3))
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)
APP(app(append, app(app(append, l1), l2)), l3) → APP(append, l2)
APP(app(append, app(app(cons, h), t)), l) → APP(app(append, t), l)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l2)
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
APP(app(append, app(app(cons, h), t)), l) → APP(app(cons, h), app(app(append, t), l))
APP(app(map, f), app(app(cons, h), t)) → APP(cons, app(f, h))

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

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(append, app(app(append, l1), l2)), l3) → APP(app(append, l2), l3)
APP(app(map, f), app(app(cons, h), t)) → APP(app(cons, app(f, h)), app(app(map, f), t))
APP(app(append, app(app(cons, h), t)), l) → APP(append, t)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(append, app(app(map, f), l1)), app(app(map, f), l2))
APP(app(map, f), app(app(append, l1), l2)) → APP(append, app(app(map, f), l1))
APP(app(append, app(app(append, l1), l2)), l3) → APP(app(append, l1), app(app(append, l2), l3))
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)
APP(app(append, app(app(append, l1), l2)), l3) → APP(append, l2)
APP(app(append, app(app(cons, h), t)), l) → APP(app(append, t), l)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l2)
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
APP(app(append, app(app(cons, h), t)), l) → APP(app(cons, h), app(app(append, t), l))
APP(app(map, f), app(app(cons, h), t)) → APP(cons, app(f, h))

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

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(append, app(app(append, l1), l2)), l3) → APP(app(append, l2), l3)
APP(app(map, f), app(app(cons, h), t)) → APP(app(cons, app(f, h)), app(app(map, f), t))
APP(app(map, f), app(app(append, l1), l2)) → APP(app(append, app(app(map, f), l1)), app(app(map, f), l2))
APP(app(append, app(app(cons, h), t)), l) → APP(append, t)
APP(app(map, f), app(app(append, l1), l2)) → APP(append, app(app(map, f), l1))
APP(app(append, app(app(append, l1), l2)), l3) → APP(app(append, l1), app(app(append, l2), l3))
APP(app(append, app(app(append, l1), l2)), l3) → APP(append, l2)
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)
APP(app(append, app(app(cons, h), t)), l) → APP(app(append, t), l)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l2)
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
APP(app(append, app(app(cons, h), t)), l) → APP(app(cons, h), app(app(append, t), l))
APP(app(map, f), app(app(cons, h), t)) → APP(cons, app(f, h))

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

APP(app(append, app(app(append, l1), l2)), l3) → APP(app(append, l2), l3)
APP(app(append, app(app(append, l1), l2)), l3) → APP(app(append, l1), app(app(append, l2), l3))
APP(app(append, app(app(cons, h), t)), l) → APP(app(append, t), l)

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

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:

APPEND(cons(h, t), l) → APPEND(t, l)
APPEND(append(l1, l2), l3) → APPEND(l2, l3)
APPEND(append(l1, l2), l3) → APPEND(l1, append(l2, l3))

The TRS R consists of the following rules:

append(nil, l) → l
append(cons(h, t), l) → cons(h, append(t, l))
append(append(l1, l2), l3) → append(l1, append(l2, l3))

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(append, app(app(append, l1), l2)), l3) → APP(app(append, l2), l3)
APP(app(append, app(app(append, l1), l2)), l3) → APP(app(append, l1), app(app(append, l2), l3))
The remaining pairs can at least be oriented weakly.

APP(app(append, app(app(cons, h), t)), l) → APP(app(append, t), l)
Used ordering: Combined order from the following AFS and order.
APPEND(x1, x2)  =  x1
cons(x1, x2)  =  x2
append(x1, x2)  =  append(x1, x2)
nil  =  nil

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(append, app(app(cons, h), t)), l) → APP(app(append, t), l)

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

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:

APPEND(cons(h, t), l) → APPEND(t, l)

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(append, app(app(cons, h), t)), l) → APP(app(append, t), l)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
APPEND(x1, x2)  =  APPEND(x1)
cons(x1, x2)  =  cons(x1, x2)

Lexicographic Path Order [19].
Precedence:
cons2 > APPEND1


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(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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

APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l2)
APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(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 oriented strictly and are deleted.


APP(app(map, f), app(app(cons, h), t)) → APP(f, h)
The remaining pairs can at least be oriented weakly.

APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l2)
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x1)
app(x1, x2)  =  app(x2)
map  =  map
append  =  append
cons  =  cons

Lexicographic Path Order [19].
Precedence:
append > APP1 > [app1, map]


The following usable rules [14] were oriented: none



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

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

APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l2)
APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

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:

MAP(f, cons(h, t)) → MAP(f, t)
MAP(f, append(l1, l2)) → MAP(f, l2)
MAP(f, append(l1, l2)) → MAP(f, l1)

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(map, f), app(app(append, l1), l2)) → APP(app(map, f), l1)
APP(app(map, f), app(app(append, l1), l2)) → APP(app(map, f), l2)
The remaining pairs can at least be oriented weakly.

APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)
Used ordering: Combined order from the following AFS and order.
MAP(x1, x2)  =  MAP(x2)
cons(x1, x2)  =  x2
append(x1, x2)  =  append(x1, x2)

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
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof

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

APP(app(map, f), app(app(cons, h), t)) → APP(app(map, f), t)

The TRS R consists of the following rules:

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(map, f), l2))

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:

MAP(f, cons(h, t)) → MAP(f, t)

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(map, f), app(app(cons, h), t)) → APP(app(map, f), t)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MAP(x1, x2)  =  MAP(x2)
cons(x1, x2)  =  cons(x2)

Lexicographic Path Order [19].
Precedence:
cons1 > MAP1


The following usable rules [14] were oriented: none



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

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

app(app(append, nil), l) → l
app(app(append, app(app(cons, h), t)), l) → app(app(cons, h), app(app(append, t), l))
app(app(map, f), nil) → nil
app(app(map, f), app(app(cons, h), t)) → app(app(cons, app(f, h)), app(app(map, f), t))
app(app(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))
app(app(map, f), app(app(append, l1), l2)) → app(app(append, app(app(map, f), l1)), app(app(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.