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:

rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))

The set Q consists of the following terms:

rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

REV1(X, cons(Y, L)) → REV1(Y, L)
REV(cons(X, L)) → REV1(X, L)
REV(cons(X, L)) → REV2(X, L)
REV2(X, cons(Y, L)) → REV(cons(X, rev(rev2(Y, L))))
REV2(X, cons(Y, L)) → REV(rev2(Y, L))
REV2(X, cons(Y, L)) → REV2(Y, L)

The TRS R consists of the following rules:

rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))

The set Q consists of the following terms:

rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

REV1(X, cons(Y, L)) → REV1(Y, L)

The TRS R consists of the following rules:

rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))

The set Q consists of the following terms:

rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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.

REV1(X, cons(Y, L)) → REV1(Y, L)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
REV1(x1, x2)  =  REV1(x2)
cons(x1, x2)  =  cons(x2)

Recursive Path Order [RPO].
Precedence:

cons₁ > REV1₁

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))

The set Q consists of the following terms:

rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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(cons(X, L)) → REV2(X, L)
REV2(X, cons(Y, L)) → REV(cons(X, rev(rev2(Y, L))))
REV2(X, cons(Y, L)) → REV(rev2(Y, L))
REV2(X, cons(Y, L)) → REV2(Y, L)

The TRS R consists of the following rules:

rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))

The set Q consists of the following terms:

rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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(cons(X, L)) → REV2(X, L)
REV2(X, cons(Y, L)) → REV(cons(X, rev(rev2(Y, L))))
REV2(X, cons(Y, L)) → REV(rev2(Y, L))
REV2(X, cons(Y, L)) → REV2(Y, L)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
REV(x1)  =  x1
cons(x1, x2)  =  cons(x2)
REV2(x1, x2)  =  REV2(x2)
rev(x1)  =  x1
rev2(x1, x2)  =  x2
rev1(x1, x2)  =  x2
s(x1)  =  x1
nil  =  nil
0  =  0

Recursive Path Order [RPO].
Precedence:

cons₁ > REV2₁
nil > REV2₁
0 > REV2₁

The following usable rules [FROCOS05] were oriented:

rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))

(14) Obligation:

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

rev1(0, nil) → 0
rev1(s(X), nil) → s(X)
rev1(X, cons(Y, L)) → rev1(Y, L)
rev(nil) → nil
rev(cons(X, L)) → cons(rev1(X, L), rev2(X, L))
rev2(X, nil) → nil
rev2(X, cons(Y, L)) → rev(cons(X, rev(rev2(Y, L))))

The set Q consists of the following terms:

rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
rev(nil)
rev(cons(x0, x1))
rev2(x0, nil)
rev2(x0, cons(x1, x2))

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