Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 PisEmptyProof (⇔)
    ↳20 TRUE

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 7 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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)
APP(app(append, app(app(append, l1), l2)), l3) → APP(app(append, l2), l3)

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.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

append1(append(l1, l2), l3) → append1(l1, append(l2, l3))
append1(cons(h, t), l) → append1(t, l)
append1(append(l1, l2), l3) → append1(l2, l3)

The a-transformed usable rules are

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


The following pairs can be oriented strictly and are deleted.


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(app(append, l2), l3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
append1(x1, x2)  =  x1
append(x1, x2)  =  append(x1, x2)
cons(x1, x2)  =  x2
nil  =  nil

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented:

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(append, app(app(append, l1), l2)), l3) → app(app(append, l1), app(app(append, l2), l3))

(7) Obligation:

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.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

append1(cons(h, t), l) → append1(t, l)

The a-transformed usable rules are
none


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.
Used ordering: Combined order from the following AFS and order.
append1(x1, x2)  =  append1(x1, x2)
cons(x1, x2)  =  cons(x2)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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.

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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

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


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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)
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 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.

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

map1(f, cons(h, t)) → map1(f, t)
map1(f, append(l1, l2)) → map1(f, l1)
map1(f, append(l1, l2)) → map1(f, l2)

The a-transformed usable rules are
none


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.
Used ordering: Combined order from the following AFS and order.
map1(x1, x2)  =  x2
cons(x1, x2)  =  x2
append(x1, x2)  =  append(x1, x2)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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.

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04]. Here, we combined the reduction pair processor with the A-transformation [FROCOS05] which results in the following intermediate Q-DP Problem.
The a-transformed P is

map1(f, cons(h, t)) → map1(f, t)

The a-transformed usable rules are
none


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.
Used ordering: Combined order from the following AFS and order.
map1(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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.

(19) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(20) TRUE