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(.(x, y)) → ++(rev(y), .(x, nil))
car(.(x, y)) → x
cdr(.(x, y)) → y
null(nil) → true
null(.(x, y)) → false
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

car(.(x, y)) → x
cdr(.(x, y)) → y
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))
rev(nil) → nil
rev(.(x, y)) → ++(rev(y), .(x, nil))

The TRS R 2 is

null(nil) → true
null(.(x, y)) → false

The signature Sigma is {null, true, false}

(2) Obligation:

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

rev(nil) → nil
rev(.(x, y)) → ++(rev(y), .(x, nil))
car(.(x, y)) → x
cdr(.(x, y)) → y
null(nil) → true
null(.(x, y)) → false
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))

The set Q consists of the following terms:

rev(nil)
rev(.(x0, x1))
car(.(x0, x1))
cdr(.(x0, x1))
null(nil)
null(.(x0, x1))
++(nil, x0)
++(.(x0, 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(.(x, y)) → ++¹(rev(y), .(x, nil))
REV(.(x, y)) → REV(y)
++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(.(x, y)) → ++(rev(y), .(x, nil))
car(.(x, y)) → x
cdr(.(x, y)) → y
null(nil) → true
null(.(x, y)) → false
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))

The set Q consists of the following terms:

rev(nil)
rev(.(x0, x1))
car(.(x0, x1))
cdr(.(x0, x1))
null(nil)
null(.(x0, x1))
++(nil, x0)
++(.(x0, 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:

++¹(.(x, y), z) → ++¹(y, z)

The TRS R consists of the following rules:

rev(nil) → nil
rev(.(x, y)) → ++(rev(y), .(x, nil))
car(.(x, y)) → x
cdr(.(x, y)) → y
null(nil) → true
null(.(x, y)) → false
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))

The set Q consists of the following terms:

rev(nil)
rev(.(x0, x1))
car(.(x0, x1))
cdr(.(x0, x1))
null(nil)
null(.(x0, x1))
++(nil, x0)
++(.(x0, 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.

++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)  =  x1
.(x1, x2)  =  .(x1, x2)
rev(x1)  =  rev(x1)
nil  =  nil
++(x1, x2)  =  ++(x1, x2)
car(x1)  =  car(x1)
cdr(x1)  =  cdr(x1)
null(x1)  =  null(x1)
true  =  true
false  =  false

Recursive Path Order [RPO].
Precedence:

rev₁ > nil
rev₁ > ++₂ > .₂
null₁ > true
null₁ > false

The following usable rules [FROCOS05] were oriented:

rev(nil) → nil
rev(.(x, y)) → ++(rev(y), .(x, nil))
car(.(x, y)) → x
cdr(.(x, y)) → y
null(nil) → true
null(.(x, y)) → false
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))

(9) Obligation:

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

rev(nil) → nil
rev(.(x, y)) → ++(rev(y), .(x, nil))
car(.(x, y)) → x
cdr(.(x, y)) → y
null(nil) → true
null(.(x, y)) → false
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))

The set Q consists of the following terms:

rev(nil)
rev(.(x0, x1))
car(.(x0, x1))
cdr(.(x0, x1))
null(nil)
null(.(x0, x1))
++(nil, x0)
++(.(x0, 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(.(x, y)) → REV(y)

The TRS R consists of the following rules:

rev(nil) → nil
rev(.(x, y)) → ++(rev(y), .(x, nil))
car(.(x, y)) → x
cdr(.(x, y)) → y
null(nil) → true
null(.(x, y)) → false
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))

The set Q consists of the following terms:

rev(nil)
rev(.(x0, x1))
car(.(x0, x1))
cdr(.(x0, x1))
null(nil)
null(.(x0, x1))
++(nil, x0)
++(.(x0, 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(.(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)  =  ++(x1, x2)
car(x1)  =  x1
cdr(x1)  =  x1
null(x1)  =  null(x1)
true  =  true
false  =  false

Recursive Path Order [RPO].
Precedence:

rev₁ > nil
rev₁ > ++₂ > .₂
null₁ > true
null₁ > false

The following usable rules [FROCOS05] were oriented:

rev(nil) → nil
rev(.(x, y)) → ++(rev(y), .(x, nil))
car(.(x, y)) → x
cdr(.(x, y)) → y
null(nil) → true
null(.(x, y)) → false
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))

(14) Obligation:

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

rev(nil) → nil
rev(.(x, y)) → ++(rev(y), .(x, nil))
car(.(x, y)) → x
cdr(.(x, y)) → y
null(nil) → true
null(.(x, y)) → false
++(nil, y) → y
++(.(x, y), z) → .(x, ++(y, z))

The set Q consists of the following terms:

rev(nil)
rev(.(x0, x1))
car(.(x0, x1))
cdr(.(x0, x1))
null(nil)
null(.(x0, x1))
++(nil, x0)
++(.(x0, 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