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(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

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(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

The set Q consists of the following terms:

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

G(cons(x, k), d) → G(k, cons(x, d))
F(a, empty) → G(a, empty)
F(a, cons(x, k)) → F(cons(x, a), k)

The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

The set Q consists of the following terms:

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

G(cons(x, k), d) → G(k, cons(x, d))
F(a, empty) → G(a, empty)
F(a, cons(x, k)) → F(cons(x, a), k)

The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

The set Q consists of the following terms:

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

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

G(cons(x, k), d) → G(k, cons(x, d))
F(a, empty) → G(a, empty)
F(a, cons(x, k)) → F(cons(x, a), k)

The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

The set Q consists of the following terms:

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

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

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

G(cons(x, k), d) → G(k, cons(x, d))

The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

The set Q consists of the following terms:

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


G(cons(x, k), d) → G(k, cons(x, d))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
G(x1, x2)  =  x1
cons(x1, x2)  =  cons(x2)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



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

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

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

The set Q consists of the following terms:

f(x0, empty)
f(x0, cons(x1, x2))
g(empty, x0)
g(cons(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof

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

F(a, cons(x, k)) → F(cons(x, a), k)

The TRS R consists of the following rules:

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

The set Q consists of the following terms:

f(x0, empty)
f(x0, cons(x1, x2))
g(empty, x0)
g(cons(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(a, cons(x, k)) → F(cons(x, a), k)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



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

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

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

The set Q consists of the following terms:

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