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(ls) -> r12(ls, empty)
r12(empty, a) -> a
r12(cons2(x, k), a) -> r12(k, cons2(x, a))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

rev1(ls) -> r12(ls, empty)
r12(empty, a) -> a
r12(cons2(x, k), a) -> r12(k, cons2(x, a))

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(ls) -> r12(ls, empty)
r12(empty, a) -> a
r12(cons2(x, k), a) -> r12(k, cons2(x, a))

The set Q consists of the following terms:

rev1(x0)
r12(empty, x0)
r12(cons2(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:

R12(cons2(x, k), a) -> R12(k, cons2(x, a))
REV1(ls) -> R12(ls, empty)

The TRS R consists of the following rules:

rev1(ls) -> r12(ls, empty)
r12(empty, a) -> a
r12(cons2(x, k), a) -> r12(k, cons2(x, a))

The set Q consists of the following terms:

rev1(x0)
r12(empty, x0)
r12(cons2(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:

R12(cons2(x, k), a) -> R12(k, cons2(x, a))
REV1(ls) -> R12(ls, empty)

The TRS R consists of the following rules:

rev1(ls) -> r12(ls, empty)
r12(empty, a) -> a
r12(cons2(x, k), a) -> r12(k, cons2(x, a))

The set Q consists of the following terms:

rev1(x0)
r12(empty, x0)
r12(cons2(x0, x1), x2)

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

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

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

R12(cons2(x, k), a) -> R12(k, cons2(x, a))

The TRS R consists of the following rules:

rev1(ls) -> r12(ls, empty)
r12(empty, a) -> a
r12(cons2(x, k), a) -> r12(k, cons2(x, a))

The set Q consists of the following terms:

rev1(x0)
r12(empty, x0)
r12(cons2(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.


R12(cons2(x, k), a) -> R12(k, cons2(x, a))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
R12(x1, x2)  =  x1
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

rev1(ls) -> r12(ls, empty)
r12(empty, a) -> a
r12(cons2(x, k), a) -> r12(k, cons2(x, a))

The set Q consists of the following terms:

rev1(x0)
r12(empty, x0)
r12(cons2(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.