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
  ↳ 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))

Q is empty.

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))

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

↳ 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))

Q is empty.
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
  ↳ 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))

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


The following pairs can be oriented strictly and are deleted.


++12(.2(x, y), z) -> ++12(y, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( .2(x1, x2) ) = 2x1 + x2 + 2


POL( ++12(x1, x2) ) = x1 + 2x2 + 1



The following usable rules [14] were oriented: none



↳ 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))

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

↳ 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))

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


The following pairs can be oriented strictly and are deleted.


REV1(.2(x, y)) -> REV1(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( .2(x1, x2) ) = 2x1 + x2 + 2


POL( REV1(x1) ) = max{0, 2x1 - 1}



The following usable rules [14] were oriented: none



↳ 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))

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