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:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, 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:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

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

REV(cons(x, l)) → REV1(x, l)
REV(cons(x, l)) → REV2(x, l)
REV1(x, cons(y, l)) → REV1(y, l)
REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))
REV2(x, cons(y, l)) → REV2(y, l)

The TRS R consists of the following rules:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

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

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

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

Lexicographic Path Order [LPO].
Precedence:

rev1 > s > [cons₁, nil, 0]

The following usable rules [FROCOS05] were oriented:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

(9) Obligation:

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

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

rev(nil)
rev(cons(x0, x1))
rev1(0, nil)
rev1(s(x0), nil)
rev1(x0, cons(x1, x2))
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, rev2(y, l)))
REV2(x, cons(y, l)) → REV2(y, l)

The TRS R consists of the following rules:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

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

Lexicographic Path Order [LPO].
Precedence:

[rev1, s] > [nil, 0]

The following usable rules [FROCOS05] were oriented:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

(14) Obligation:

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

REV2(x, cons(y, l)) → REV(cons(x, rev2(y, l)))

The TRS R consists of the following rules:

rev(nil) → nil
rev(cons(x, l)) → cons(rev1(x, l), rev2(x, l))
rev1(0, nil) → 0
rev1(s(x), nil) → s(x)
rev1(x, cons(y, l)) → rev1(y, l)
rev2(x, nil) → nil
rev2(x, cons(y, l)) → rev(cons(x, rev2(y, l)))

The set Q consists of the following terms:

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

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

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(16) TRUE