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 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 PisEmptyProof (⇔)
        ↳14 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].


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(cons, h), t)), l) → APP(app(append, t), l)
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: Lexicographic path order with status [LPO].
Quasi-Precedence:
[APP2, app2, append, cons]

Status:
APP2: [1,2]
app2: [1,2]
append: []
cons: []
nil: []


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:
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.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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.

(11) 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(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 remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
[APP2, map] > [app2, append]
cons > [app2, append]

Status:
APP2: [2,1]
app2: [2,1]
map: []
cons: []
append: []


The following usable rules [FROCOS05] were oriented: none

(12) 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.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE