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:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

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:

FLATTEN(unit(x)) → FLATTEN(x)
FLATTEN(++(x, y)) → ++¹(flatten(x), flatten(y))
FLATTEN(++(x, y)) → FLATTEN(x)
FLATTEN(++(x, y)) → FLATTEN(y)
FLATTEN(++(unit(x), y)) → ++¹(flatten(x), flatten(y))
FLATTEN(++(unit(x), y)) → FLATTEN(x)
FLATTEN(++(unit(x), y)) → FLATTEN(y)
REV(++(x, y)) → ++¹(rev(y), rev(x))
REV(++(x, y)) → REV(y)
REV(++(x, y)) → REV(x)
++¹(++(x, y), z) → ++¹(x, ++(y, z))
++¹(++(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

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 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

++¹(++(x, y), z) → ++¹(y, z)
++¹(++(x, y), z) → ++¹(x, ++(y, z))

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

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.

++¹(++(x, y), z) → ++¹(y, z)
++¹(++(x, y), z) → ++¹(x, ++(y, z))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
++¹(x1, x2)  =  x1
++(x1, x2)  =  ++(x1, x2)
nil  =  nil

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

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:

REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

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.

REV(++(x, y)) → REV(x)
REV(++(x, y)) → REV(y)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

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:

FLATTEN(++(x, y)) → FLATTEN(x)
FLATTEN(unit(x)) → FLATTEN(x)
FLATTEN(++(x, y)) → FLATTEN(y)
FLATTEN(++(unit(x), y)) → FLATTEN(x)
FLATTEN(++(unit(x), y)) → FLATTEN(y)

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

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.

FLATTEN(++(x, y)) → FLATTEN(x)
FLATTEN(++(x, y)) → FLATTEN(y)
FLATTEN(++(unit(x), y)) → FLATTEN(x)
FLATTEN(++(unit(x), y)) → FLATTEN(y)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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

FLATTEN(unit(x)) → FLATTEN(x)

The TRS R consists of the following rules:

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

Q is empty.
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.

FLATTEN(unit(x)) → FLATTEN(x)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

flatten(nil) → nil
flatten(unit(x)) → flatten(x)
flatten(++(x, y)) → ++(flatten(x), flatten(y))
flatten(++(unit(x), y)) → ++(flatten(x), flatten(y))
flatten(flatten(x)) → flatten(x)
rev(nil) → nil
rev(unit(x)) → unit(x)
rev(++(x, y)) → ++(rev(y), rev(x))
rev(rev(x)) → x
++(x, nil) → x
++(nil, y) → y
++(++(x, y), z) → ++(x, ++(y, z))

Q is empty.
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