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(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: Combined order from the following AFS and order.
++¹(x1, x2)  =  ++¹(x2)
++(x1, x2)  =  ++(x1, x2)
.(x1, x2)  =  .
rev(x1)  =  rev(x1)
nil  =  nil
make(x1)  =  x1

Recursive path order with status [RPO].
Precedence:

++^1₁ > ++₂ > .
rev₁ > ++₂ > .
rev₁ > nil > .

Status:

++^1₁: [1]
++₂: [2,1]
.: multiset
rev₁: [1]
nil: multiset

The following usable rules [FROCOS05] were oriented:

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)

(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: Combined order from the following AFS and order.
++¹(x1, x2)  =  ++¹(x1)
.(x1, x2)  =  .(x2)
rev(x1)  =  rev(x1)
nil  =  nil
++(x1, x2)  =  ++(x1, x2)
make(x1)  =  make(x1)

Recursive path order with status [RPO].
Precedence:

++^1₁ > .₁
rev₁ > ++₂ > .₁
make₁ > nil > .₁

Status:

++^1₁: multiset
.₁: [1]
rev₁: [1]
nil: multiset
++₂: [2,1]
make₁: multiset

The following usable rules [FROCOS05] were oriented:

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)

(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: Combined order from the following AFS and order.
REV(x1)  =  x1
++(x1, x2)  =  ++(x1, x2)
rev(x1)  =  rev(x1)
nil  =  nil
.(x1, x2)  =  .
make(x1)  =  x1

Recursive path order with status [RPO].
Precedence:

rev₁ > ++₂ > .
rev₁ > nil > .

Status:

++₂: [2,1]
rev₁: multiset
nil: multiset
.: multiset

The following usable rules [FROCOS05] were oriented:

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)

(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