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:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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:

D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)
H(e(x), y) → D(x, y)
D(g(x, y), z) → D(x, z)
D(g(g(0, x), y), s(z)) → G(e(x), d(g(g(0, x), y), z))
H(e(x), y) → H(d(x, y), s(y))
G(e(x), e(y)) → G(x, y)
D(g(x, y), z) → G(d(x, z), e(y))

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)
H(e(x), y) → D(x, y)
D(g(x, y), z) → D(x, z)
D(g(g(0, x), y), s(z)) → G(e(x), d(g(g(0, x), y), z))
H(e(x), y) → H(d(x, y), s(y))
G(e(x), e(y)) → G(x, y)
D(g(x, y), z) → G(d(x, z), e(y))

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)
H(e(x), y) → D(x, y)
D(g(g(0, x), y), s(z)) → G(e(x), d(g(g(0, x), y), z))
D(g(x, y), z) → D(x, z)
H(e(x), y) → H(d(x, y), s(y))
G(e(x), e(y)) → G(x, y)
D(g(x, y), z) → G(d(x, z), e(y))

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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

G(e(x), e(y)) → G(x, y)

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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.


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

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP

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

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP

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

D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)
D(g(x, y), z) → D(x, z)

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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.


D(g(x, y), z) → D(x, z)
The remaining pairs can at least be oriented weakly.

D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)
Used ordering: Combined order from the following AFS and order.
D(x1, x2)  =  x1
g(x1, x2)  =  g(x1)
0  =  0
s(x1)  =  s

Lexicographic Path Order [19].
Precedence:
0 > g1
s > g1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof
              ↳ QDP

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

D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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.


D(g(g(0, x), y), s(z)) → D(g(g(0, x), y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
D(x1, x2)  =  D(x1, x2)
g(x1, x2)  =  g(x1, x2)
0  =  0
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
[D2, 0] > g2


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
              ↳ QDP

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

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP

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

H(e(x), y) → H(d(x, y), s(y))

The TRS R consists of the following rules:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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