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:

f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))

The set Q consists of the following terms:

f(empty, x0)
f(cons(x0, x1), x2)
g(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:

F(cons(x, k), l) → G(k, l, cons(x, k))
G(a, b, c) → F(a, cons(b, c))

The TRS R consists of the following rules:

f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))

The set Q consists of the following terms:

f(empty, x0)
f(cons(x0, x1), x2)
g(x0, x1, x2)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

F(cons(x, k), l) → G(k, l, cons(x, k))
G(a, b, c) → F(a, cons(b, c))

The TRS R consists of the following rules:

f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))

The set Q consists of the following terms:

f(empty, x0)
f(cons(x0, x1), x2)
g(x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ QDPOrderProof

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

F(cons(x, k), l) → G(k, l, cons(x, k))
G(a, b, c) → F(a, cons(b, c))

The TRS R consists of the following rules:

f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))

The set Q consists of the following terms:

f(empty, x0)
f(cons(x0, x1), x2)
g(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 oriented strictly and are deleted.


F(cons(x, k), l) → G(k, l, cons(x, k))
The remaining pairs can at least be oriented weakly.

G(a, b, c) → F(a, cons(b, c))
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  F(x1)
cons(x1, x2)  =  cons(x2)
G(x1, x2, x3)  =  G(x1)

Recursive Path Order [2].
Precedence:
cons1 > [F1, G1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ DependencyGraphProof

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

G(a, b, c) → F(a, cons(b, c))

The TRS R consists of the following rules:

f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))

The set Q consists of the following terms:

f(empty, x0)
f(cons(x0, x1), x2)
g(x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.