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

(0) Obligation:

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

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(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

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(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

APP(app(map, f), app(app(cons, x), xs)) → APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) → APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(flatten, app(app(node, x), xs)) → APP(app(cons, x), app(concat, app(app(map, flatten), xs)))
APP(flatten, app(app(node, x), xs)) → APP(cons, x)
APP(flatten, app(app(node, x), xs)) → APP(concat, app(app(map, flatten), xs))
APP(flatten, app(app(node, x), xs)) → APP(app(map, flatten), xs)
APP(flatten, app(app(node, x), xs)) → APP(map, flatten)
APP(concat, app(app(cons, x), xs)) → APP(app(append, x), app(concat, xs))
APP(concat, app(app(cons, x), xs)) → APP(append, x)
APP(concat, app(app(cons, x), xs)) → APP(concat, xs)
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(cons, x), app(app(append, xs), ys))
APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)
APP(app(append, app(app(cons, x), xs)), ys) → APP(append, xs)

The TRS R consists of the following rules:

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(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 10 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)

The TRS R consists of the following rules:

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(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)

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(x, xs), ys) → append1(xs, ys)

The a-transformed usable rules are
none

The following pairs can be oriented strictly and are deleted.

APP(app(append, app(app(cons, x), xs)), ys) → APP(app(append, xs), ys)

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

Lexicographic path order with status [LPO].
Precedence:

cons₂ > append1₁

Status:

cons₂: [1,2]
append1₁: [1]

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(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(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)

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(concat, app(app(cons, x), xs)) → APP(concat, xs)

The TRS R consists of the following rules:

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(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)

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

(13) 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

concat1(cons(x, xs)) → concat1(xs)

The a-transformed usable rules are
none

The following pairs can be oriented strictly and are deleted.

APP(concat, app(app(cons, x), xs)) → APP(concat, xs)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
concat1(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

cons₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

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(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

APP(app(map, f), app(app(cons, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(flatten, app(app(node, x), xs)) → APP(app(map, flatten), xs)

The TRS R consists of the following rules:

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(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)

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

(18) 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, x), xs)) → APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) → APP(f, x)
APP(flatten, app(app(node, x), xs)) → APP(app(map, flatten), xs)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  APP(x2)
app(x1, x2)  =  app(x1, x2)
map  =  map
cons  =  cons
flatten  =  flatten
node  =  node

Lexicographic path order with status [LPO].
Precedence:

map > APP₁ > flatten > app₂

Status:

APP₁: [1]
cons: []
map: []
node: []
app₂: [1,2]
flatten: []

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

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(flatten, app(app(node, x), xs)) → app(app(cons, x), app(concat, app(app(map, flatten), xs)))
app(concat, nil) → nil
app(concat, app(app(cons, x), xs)) → app(app(append, x), app(concat, xs))
app(app(append, nil), xs) → xs
app(app(append, app(app(cons, x), xs)), ys) → app(app(cons, x), app(app(append, xs), ys))

The set Q consists of the following terms:

app(app(map, x0), nil)
app(app(map, x0), app(app(cons, x1), x2))
app(flatten, app(app(node, x0), x1))
app(concat, nil)
app(concat, app(app(cons, x0), x1))
app(app(append, nil), x0)
app(app(append, app(app(cons, x0), x1)), x2)

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE