Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

rev1(nil) -> nil
rev1(.2(x, y)) -> ++2(rev1(y), .2(x, nil))
car1(.2(x, y)) -> x
cdr1(.2(x, y)) -> y
null1(nil) -> true
null1(.2(x, y)) -> false
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

rev1(nil) -> nil
rev1(.2(x, y)) -> ++2(rev1(y), .2(x, nil))
car1(.2(x, y)) -> x
cdr1(.2(x, y)) -> y
null1(nil) -> true
null1(.2(x, y)) -> false
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

rev1(nil) -> nil
rev1(.2(x, y)) -> ++2(rev1(y), .2(x, nil))
car1(.2(x, y)) -> x
cdr1(.2(x, y)) -> y
null1(nil) -> true
null1(.2(x, y)) -> false
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))

The set Q consists of the following terms:

rev1(nil)
rev1(.2(x0, x1))
car1(.2(x0, x1))
cdr1(.2(x0, x1))
null1(nil)
null1(.2(x0, x1))
++2(nil, x0)
++2(.2(x0, x1), x2)


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

REV1(.2(x, y)) -> ++12(rev1(y), .2(x, nil))
REV1(.2(x, y)) -> REV1(y)
++12(.2(x, y), z) -> ++12(y, z)

The TRS R consists of the following rules:

rev1(nil) -> nil
rev1(.2(x, y)) -> ++2(rev1(y), .2(x, nil))
car1(.2(x, y)) -> x
cdr1(.2(x, y)) -> y
null1(nil) -> true
null1(.2(x, y)) -> false
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))

The set Q consists of the following terms:

rev1(nil)
rev1(.2(x0, x1))
car1(.2(x0, x1))
cdr1(.2(x0, x1))
null1(nil)
null1(.2(x0, x1))
++2(nil, x0)
++2(.2(x0, x1), x2)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

REV1(.2(x, y)) -> ++12(rev1(y), .2(x, nil))
REV1(.2(x, y)) -> REV1(y)
++12(.2(x, y), z) -> ++12(y, z)

The TRS R consists of the following rules:

rev1(nil) -> nil
rev1(.2(x, y)) -> ++2(rev1(y), .2(x, nil))
car1(.2(x, y)) -> x
cdr1(.2(x, y)) -> y
null1(nil) -> true
null1(.2(x, y)) -> false
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))

The set Q consists of the following terms:

rev1(nil)
rev1(.2(x0, x1))
car1(.2(x0, x1))
cdr1(.2(x0, x1))
null1(nil)
null1(.2(x0, x1))
++2(nil, x0)
++2(.2(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

++12(.2(x, y), z) -> ++12(y, z)

The TRS R consists of the following rules:

rev1(nil) -> nil
rev1(.2(x, y)) -> ++2(rev1(y), .2(x, nil))
car1(.2(x, y)) -> x
cdr1(.2(x, y)) -> y
null1(nil) -> true
null1(.2(x, y)) -> false
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))

The set Q consists of the following terms:

rev1(nil)
rev1(.2(x0, x1))
car1(.2(x0, x1))
cdr1(.2(x0, x1))
null1(nil)
null1(.2(x0, x1))
++2(nil, x0)
++2(.2(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


++12(.2(x, y), z) -> ++12(y, z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
++12(x1, x2)  =  ++11(x1)
.2(x1, x2)  =  .1(x2)

Lexicographic Path Order [19].
Precedence:
[++^11, .1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

rev1(nil) -> nil
rev1(.2(x, y)) -> ++2(rev1(y), .2(x, nil))
car1(.2(x, y)) -> x
cdr1(.2(x, y)) -> y
null1(nil) -> true
null1(.2(x, y)) -> false
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))

The set Q consists of the following terms:

rev1(nil)
rev1(.2(x0, x1))
car1(.2(x0, x1))
cdr1(.2(x0, x1))
null1(nil)
null1(.2(x0, x1))
++2(nil, x0)
++2(.2(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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

REV1(.2(x, y)) -> REV1(y)

The TRS R consists of the following rules:

rev1(nil) -> nil
rev1(.2(x, y)) -> ++2(rev1(y), .2(x, nil))
car1(.2(x, y)) -> x
cdr1(.2(x, y)) -> y
null1(nil) -> true
null1(.2(x, y)) -> false
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))

The set Q consists of the following terms:

rev1(nil)
rev1(.2(x0, x1))
car1(.2(x0, x1))
cdr1(.2(x0, x1))
null1(nil)
null1(.2(x0, x1))
++2(nil, x0)
++2(.2(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


REV1(.2(x, y)) -> REV1(y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
REV1(x1)  =  REV1(x1)
.2(x1, x2)  =  .2(x1, x2)

Lexicographic Path Order [19].
Precedence:
.2 > REV1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof

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

rev1(nil) -> nil
rev1(.2(x, y)) -> ++2(rev1(y), .2(x, nil))
car1(.2(x, y)) -> x
cdr1(.2(x, y)) -> y
null1(nil) -> true
null1(.2(x, y)) -> false
++2(nil, y) -> y
++2(.2(x, y), z) -> .2(x, ++2(y, z))

The set Q consists of the following terms:

rev1(nil)
rev1(.2(x0, x1))
car1(.2(x0, x1))
cdr1(.2(x0, x1))
null1(nil)
null1(.2(x0, x1))
++2(nil, x0)
++2(.2(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.