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)


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 contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

++12(.2(x, y), z) -> ++12(y, z)
Used argument filtering: ++12(x1, x2)  =  x1
.2(x1, x2)  =  .1(x2)
Used ordering: Precedence:
trivial



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

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

REV1(.2(x, y)) -> REV1(y)
Used argument filtering: REV1(x1)  =  x1
.2(x1, x2)  =  .1(x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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.