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
  ↳ 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))

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:

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))

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:

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))

Q is empty.
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
  ↳ 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))

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.


R12(cons2(x, k), a) -> R12(k, cons2(x, a))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( R12(x1, x2) ) = x1


POL( cons2(x1, x2) ) = x2 + 1



The following usable rules [14] were oriented: none



↳ 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))

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.