(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) 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(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))

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 3 SCCs with 10 less nodes.

(4) Complex Obligation (AND)

(5) 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))

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(cons, x), xs)), ys) → APP(app(append, xs), ys)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(APP(x1, x2)) = x1   
POL(app(x1, x2)) = 1 + x1 + x2   
POL(append) = 0   
POL(cons) = 0   

The following usable rules [FROCOS05] were oriented: none

(7) 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))

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(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))

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(concat, app(app(cons, x), xs)) → APP(concat, xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(APP(x1, x2)) = x2   
POL(app(x1, x2)) = 1 + x1 + x2   
POL(concat) = 0   
POL(cons) = 0   

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(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.
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

(15) 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))

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

(16) 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: Polynomial interpretation [POLO]:

POL(APP(x1, x2)) = x2   
POL(app(x1, x2)) = x1 + x2   
POL(cons) = 1   
POL(flatten) = 0   
POL(map) = 0   
POL(node) = 1   

The following usable rules [FROCOS05] were oriented: none

(17) 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))

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE