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) 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), .(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))

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(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))

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

(6) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
.(x1, x2)  =  .(x2)

From the DPs we obtained the following set of size-change graphs:

• ++¹(.(x, y), z) → ++¹(y, z) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1, 2 >= 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) 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))

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
.(x1, x2)  =  .(x2)

From the DPs we obtained the following set of size-change graphs:

• REV(.(x, y)) → REV(y) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE