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

(0) Obligation:

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

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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:

REV(++(x, y)) → ++1(rev(y), rev(x))
REV(++(x, y)) → REV(y)
REV(++(x, y)) → REV(x)
++1(.(x, y), z) → ++1(y, z)
++1(x, ++(y, z)) → ++1(++(x, y), z)
++1(x, ++(y, z)) → ++1(x, y)

The TRS R consists of the following rules:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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

Tags:
++1 has tags [1,0]

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
++(x1, x2)  =  ++(x1, x2)
.(x1, x2)  =  x1
nil  =  nil

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

++1(.(x, y), z) → ++1(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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(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(x1, x2)  =  ++1(x1)

Tags:
++1 has tags [1,0]

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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:

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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(x1)  =  REV(x1)

Tags:
REV has tags [0]

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

rev(nil) → nil
rev(rev(x)) → x
rev(++(x, y)) → ++(rev(y), rev(x))
++(nil, y) → y
++(x, nil) → x
++(.(x, y), z) → .(x, ++(y, z))
++(x, ++(y, z)) → ++(++(x, y), z)
make(x) → .(x, nil)

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