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
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 PisEmptyProof (⇔)
    ↳19 TRUE

(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)
flatten(x1)  =  x1
nil  =  nil
unit(x1)  =  unit(x1)
rev(x1)  =  rev(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
nil > ++2
unit1 > ++2
rev1 > ++2

Status:
++2: [1,2]
nil: multiset
unit1: multiset
rev1: multiset


The following usable rules [FROCOS05] were oriented:

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))

(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: Combined order from the following AFS and order.
REV(x1)  =  x1
++(x1, x2)  =  ++(x1, x2)
flatten(x1)  =  x1
nil  =  nil
unit(x1)  =  unit(x1)
rev(x1)  =  rev(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
nil > ++2
unit1 > ++2
rev1 > ++2

Status:
++2: [1,2]
nil: multiset
unit1: multiset
rev1: multiset


The following usable rules [FROCOS05] were oriented:

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))

(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(unit(x)) → 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)  =  unit(x1)
flatten(x1)  =  x1
nil  =  nil
rev(x1)  =  rev(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
rev1 > [FLATTEN1, unit1] > ++2
rev1 > nil > ++2

Status:
FLATTEN1: multiset
++2: [1,2]
unit1: multiset
nil: multiset
rev1: multiset


The following usable rules [FROCOS05] were oriented:

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))

(17) 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.

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE