Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

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(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

The set Q consists of the following terms:

f(nil)
f(.(nil, x0))
f(.(.(x0, x1), x2))
g(nil)
g(.(x0, nil))
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(.(nil, y)) → F(y)
F(.(.(x, y), z)) → F(.(x, .(y, z)))
G(.(x, .(y, z))) → G(.(.(x, y), z))
G(.(x, nil)) → G(x)

The TRS R consists of the following rules:

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

The set Q consists of the following terms:

f(nil)
f(.(nil, x0))
f(.(.(x0, x1), x2))
g(nil)
g(.(x0, nil))
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(.(nil, y)) → F(y)
F(.(.(x, y), z)) → F(.(x, .(y, z)))
G(.(x, .(y, z))) → G(.(.(x, y), z))
G(.(x, nil)) → G(x)

The TRS R consists of the following rules:

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

The set Q consists of the following terms:

f(nil)
f(.(nil, x0))
f(.(.(x0, x1), x2))
g(nil)
g(.(x0, nil))
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
              ↳ DependencyGraphProof

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

F(.(nil, y)) → F(y)
F(.(.(x, y), z)) → F(.(x, .(y, z)))
G(.(x, .(y, z))) → G(.(.(x, y), z))
G(.(x, nil)) → G(x)

The TRS R consists of the following rules:

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

The set Q consists of the following terms:

f(nil)
f(.(nil, x0))
f(.(.(x0, x1), x2))
g(nil)
g(.(x0, nil))
g(.(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.

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

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

G(.(x, .(y, z))) → G(.(.(x, y), z))
G(.(x, nil)) → G(x)

The TRS R consists of the following rules:

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

The set Q consists of the following terms:

f(nil)
f(.(nil, x0))
f(.(.(x0, x1), x2))
g(nil)
g(.(x0, nil))
g(.(x0, .(x1, x2)))

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

↳ 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(.(nil, y)) → F(y)
F(.(.(x, y), z)) → F(.(x, .(y, z)))

The TRS R consists of the following rules:

f(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

The set Q consists of the following terms:

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

Lexicographic Path Order [19].
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(nil) → nil
f(.(nil, y)) → .(nil, f(y))
f(.(.(x, y), z)) → f(.(x, .(y, z)))
g(nil) → nil
g(.(x, nil)) → .(g(x), nil)
g(.(x, .(y, z))) → g(.(.(x, y), z))

The set Q consists of the following terms:

f(nil)
f(.(nil, x0))
f(.(.(x0, x1), x2))
g(nil)
g(.(x0, nil))
g(.(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.