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:

h2(e1(x), y) -> h2(d2(x, y), s1(y))
d2(g2(g2(0, x), y), s1(z)) -> g2(e1(x), d2(g2(g2(0, x), y), z))
d2(g2(g2(0, x), y), 0) -> e1(y)
d2(g2(0, x), y) -> e1(x)
d2(g2(x, y), z) -> g2(d2(x, z), e1(y))
g2(e1(x), e1(y)) -> e1(g2(x, y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

h2(e1(x), y) -> h2(d2(x, y), s1(y))
d2(g2(g2(0, x), y), s1(z)) -> g2(e1(x), d2(g2(g2(0, x), y), z))
d2(g2(g2(0, x), y), 0) -> e1(y)
d2(g2(0, x), y) -> e1(x)
d2(g2(x, y), z) -> g2(d2(x, z), e1(y))
g2(e1(x), e1(y)) -> e1(g2(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:

D2(g2(g2(0, x), y), s1(z)) -> D2(g2(g2(0, x), y), z)
H2(e1(x), y) -> D2(x, y)
D2(g2(x, y), z) -> D2(x, z)
D2(g2(g2(0, x), y), s1(z)) -> G2(e1(x), d2(g2(g2(0, x), y), z))
H2(e1(x), y) -> H2(d2(x, y), s1(y))
G2(e1(x), e1(y)) -> G2(x, y)
D2(g2(x, y), z) -> G2(d2(x, z), e1(y))

The TRS R consists of the following rules:

h2(e1(x), y) -> h2(d2(x, y), s1(y))
d2(g2(g2(0, x), y), s1(z)) -> g2(e1(x), d2(g2(g2(0, x), y), z))
d2(g2(g2(0, x), y), 0) -> e1(y)
d2(g2(0, x), y) -> e1(x)
d2(g2(x, y), z) -> g2(d2(x, z), e1(y))
g2(e1(x), e1(y)) -> e1(g2(x, y))

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:

D2(g2(g2(0, x), y), s1(z)) -> D2(g2(g2(0, x), y), z)
H2(e1(x), y) -> D2(x, y)
D2(g2(x, y), z) -> D2(x, z)
D2(g2(g2(0, x), y), s1(z)) -> G2(e1(x), d2(g2(g2(0, x), y), z))
H2(e1(x), y) -> H2(d2(x, y), s1(y))
G2(e1(x), e1(y)) -> G2(x, y)
D2(g2(x, y), z) -> G2(d2(x, z), e1(y))

The TRS R consists of the following rules:

h2(e1(x), y) -> h2(d2(x, y), s1(y))
d2(g2(g2(0, x), y), s1(z)) -> g2(e1(x), d2(g2(g2(0, x), y), z))
d2(g2(g2(0, x), y), 0) -> e1(y)
d2(g2(0, x), y) -> e1(x)
d2(g2(x, y), z) -> g2(d2(x, z), e1(y))
g2(e1(x), e1(y)) -> e1(g2(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

G2(e1(x), e1(y)) -> G2(x, y)

The TRS R consists of the following rules:

h2(e1(x), y) -> h2(d2(x, y), s1(y))
d2(g2(g2(0, x), y), s1(z)) -> g2(e1(x), d2(g2(g2(0, x), y), z))
d2(g2(g2(0, x), y), 0) -> e1(y)
d2(g2(0, x), y) -> e1(x)
d2(g2(x, y), z) -> g2(d2(x, z), e1(y))
g2(e1(x), e1(y)) -> e1(g2(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.


G2(e1(x), e1(y)) -> G2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(G2(x1, x2)) = x1·x2   
POL(e1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

h2(e1(x), y) -> h2(d2(x, y), s1(y))
d2(g2(g2(0, x), y), s1(z)) -> g2(e1(x), d2(g2(g2(0, x), y), z))
d2(g2(g2(0, x), y), 0) -> e1(y)
d2(g2(0, x), y) -> e1(x)
d2(g2(x, y), z) -> g2(d2(x, z), e1(y))
g2(e1(x), e1(y)) -> e1(g2(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

D2(g2(g2(0, x), y), s1(z)) -> D2(g2(g2(0, x), y), z)
D2(g2(x, y), z) -> D2(x, z)

The TRS R consists of the following rules:

h2(e1(x), y) -> h2(d2(x, y), s1(y))
d2(g2(g2(0, x), y), s1(z)) -> g2(e1(x), d2(g2(g2(0, x), y), z))
d2(g2(g2(0, x), y), 0) -> e1(y)
d2(g2(0, x), y) -> e1(x)
d2(g2(x, y), z) -> g2(d2(x, z), e1(y))
g2(e1(x), e1(y)) -> e1(g2(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.


D2(g2(x, y), z) -> D2(x, z)
The remaining pairs can at least be oriented weakly.

D2(g2(g2(0, x), y), s1(z)) -> D2(g2(g2(0, x), y), z)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(D2(x1, x2)) = x1   
POL(g2(x1, x2)) = 1 + x1   
POL(s1(x1)) = 0   

The following usable rules [14] were oriented: none



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

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

D2(g2(g2(0, x), y), s1(z)) -> D2(g2(g2(0, x), y), z)

The TRS R consists of the following rules:

h2(e1(x), y) -> h2(d2(x, y), s1(y))
d2(g2(g2(0, x), y), s1(z)) -> g2(e1(x), d2(g2(g2(0, x), y), z))
d2(g2(g2(0, x), y), 0) -> e1(y)
d2(g2(0, x), y) -> e1(x)
d2(g2(x, y), z) -> g2(d2(x, z), e1(y))
g2(e1(x), e1(y)) -> e1(g2(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.


D2(g2(g2(0, x), y), s1(z)) -> D2(g2(g2(0, x), y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(D2(x1, x2)) = x2   
POL(g2(x1, x2)) = 0   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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:

h2(e1(x), y) -> h2(d2(x, y), s1(y))
d2(g2(g2(0, x), y), s1(z)) -> g2(e1(x), d2(g2(g2(0, x), y), z))
d2(g2(g2(0, x), y), 0) -> e1(y)
d2(g2(0, x), y) -> e1(x)
d2(g2(x, y), z) -> g2(d2(x, z), e1(y))
g2(e1(x), e1(y)) -> e1(g2(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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

H2(e1(x), y) -> H2(d2(x, y), s1(y))

The TRS R consists of the following rules:

h2(e1(x), y) -> h2(d2(x, y), s1(y))
d2(g2(g2(0, x), y), s1(z)) -> g2(e1(x), d2(g2(g2(0, x), y), z))
d2(g2(g2(0, x), y), 0) -> e1(y)
d2(g2(0, x), y) -> e1(x)
d2(g2(x, y), z) -> g2(d2(x, z), e1(y))
g2(e1(x), e1(y)) -> e1(g2(x, y))

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