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

(0) Obligation:

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

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(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:

REV(++(x, y)) → REV1(x, y)
REV(++(x, y)) → REV2(x, y)
REV1(x, ++(y, z)) → REV1(y, z)
REV2(x, ++(y, z)) → REV(++(x, rev(rev2(y, z))))
REV2(x, ++(y, z)) → REV(rev2(y, z))
REV2(x, ++(y, z)) → REV2(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(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 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

REV1(x, ++(y, z)) → REV1(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(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.


REV1(x, ++(y, z)) → REV1(y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
REV1(x0, x1, x2)  =  REV1(x2)

Tags:
REV1 has argument tags [3,3,0] 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 + x2   
POL(REV1(x1, x2)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(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)) → REV2(x, y)
REV2(x, ++(y, z)) → REV(++(x, rev(rev2(y, z))))
REV2(x, ++(y, z)) → REV(rev2(y, z))
REV2(x, ++(y, z)) → REV2(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(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)) → REV2(x, y)
REV2(x, ++(y, z)) → REV(++(x, rev(rev2(y, z))))
REV2(x, ++(y, z)) → REV(rev2(y, z))
REV2(x, ++(y, z)) → REV2(y, z)
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(x0, x1)
REV2(x0, x1, x2)  =  REV2(x0)

Tags:
REV has argument tags [0,2] and root tag 0
REV2 has argument tags [2,4,6] and root tag 1

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

POL(++(x1, x2)) = 1 + x2   
POL(REV(x1)) = 1   
POL(REV2(x1, x2)) = x2   
POL(nil) = 0   
POL(rev(x1)) = x1   
POL(rev1(x1, x2)) = 0   
POL(rev2(x1, x2)) = x2   

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

rev(nil) → nil
rev(++(x, y)) → ++(rev1(x, y), rev2(x, y))
rev1(x, nil) → x
rev1(x, ++(y, z)) → rev1(y, z)
rev2(x, nil) → nil
rev2(x, ++(y, z)) → rev(++(x, rev(rev2(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