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 QDPOrderProof (⇔)
        ↳9 QDP
        ↳10 PisEmptyProof (⇔)
        ↳11 TRUE
    ↳12 QDP
        ↳13 QDPOrderProof (⇔)
        ↳14 QDP
        ↳15 PisEmptyProof (⇔)
        ↳16 TRUE
    ↳17 QDP
        ↳18 QDPOrderProof (⇔)
        ↳19 QDP
        ↳20 QDPOrderProof (⇔)
        ↳21 QDP
        ↳22 PisEmptyProof (⇔)
        ↳23 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)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
++1(x0, x1, x2)  =  ++1(x0, x1)

Tags:
++1 has argument tags [2,2,3] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(++(x1, x2)) = 1 + x1 + x2   
POL(++1(x1, x2)) = x1 + x2   
POL(nil) = 0   

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

++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.

(8) 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(x, ++(y, z))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
++1(x0, x1, x2)  =  ++1(x1)

Tags:
++1 has argument tags [3,3,3] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(++(x1, x2)) = 1 + x1   
POL(++1(x1, x2)) = 0   
POL(nil) = 0   

The following usable rules [FROCOS05] were oriented: none

(9) 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.

(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:

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.

(13) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
REV(x0, x1)  =  REV(x1)

Tags:
REV has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(++(x1, x2)) = 1 + x1 + x2   
POL(REV(x1)) = 0   

The following usable rules [FROCOS05] were oriented: none

(14) 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.

(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:

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.

(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)
FLATTEN(++(unit(x), y)) → FLATTEN(x)
FLATTEN(++(unit(x), y)) → FLATTEN(y)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
FLATTEN(x0, x1)  =  FLATTEN(x0, x1)

Tags:
FLATTEN has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(++(x1, x2)) = x1 + x2   
POL(FLATTEN(x1)) = 0   
POL(unit(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

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

(20) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
FLATTEN(x0, x1)  =  FLATTEN(x1)

Tags:
FLATTEN has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(++(x1, x2)) = 1 + x1 + x2   
POL(FLATTEN(x1)) = 0   

The following usable rules [FROCOS05] were oriented: none

(21) 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.

(22) PisEmptyProof (EQUIVALENT transformation)

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

(23) TRUE