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(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, 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(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, 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:

H2(z, e1(x)) -> H2(c1(z), d2(z, x))
D2(c1(z), g2(g2(x, y), 0)) -> G2(d2(c1(z), g2(x, y)), d2(z, g2(x, y)))
D2(z, g2(x, y)) -> D2(z, y)
H2(z, e1(x)) -> D2(z, x)
G2(e1(x), e1(y)) -> G2(x, y)
D2(c1(z), g2(g2(x, y), 0)) -> D2(z, g2(x, y))
D2(z, g2(x, y)) -> G2(e1(x), d2(z, y))
D2(c1(z), g2(g2(x, y), 0)) -> D2(c1(z), g2(x, y))

The TRS R consists of the following rules:

h2(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, 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:

H2(z, e1(x)) -> H2(c1(z), d2(z, x))
D2(c1(z), g2(g2(x, y), 0)) -> G2(d2(c1(z), g2(x, y)), d2(z, g2(x, y)))
D2(z, g2(x, y)) -> D2(z, y)
H2(z, e1(x)) -> D2(z, x)
G2(e1(x), e1(y)) -> G2(x, y)
D2(c1(z), g2(g2(x, y), 0)) -> D2(z, g2(x, y))
D2(z, g2(x, y)) -> G2(e1(x), d2(z, y))
D2(c1(z), g2(g2(x, y), 0)) -> D2(c1(z), g2(x, y))

The TRS R consists of the following rules:

h2(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, 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(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, 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)) = 3·x1 + 3·x2   
POL(e1(x1)) = 2 + 2·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(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, 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(z, g2(x, y)) -> D2(z, y)
D2(c1(z), g2(g2(x, y), 0)) -> D2(z, g2(x, y))
D2(c1(z), g2(g2(x, y), 0)) -> D2(c1(z), g2(x, y))

The TRS R consists of the following rules:

h2(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, 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(c1(z), g2(g2(x, y), 0)) -> D2(z, g2(x, y))
The remaining pairs can at least be oriented weakly.

D2(z, g2(x, y)) -> D2(z, y)
D2(c1(z), g2(g2(x, y), 0)) -> D2(c1(z), g2(x, y))
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(D2(x1, x2)) = 3·x1   
POL(c1(x1)) = 3 + 3·x1   
POL(e1(x1)) = 3   
POL(g2(x1, x2)) = 1 + 3·x1 + 2·x2   

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(z, g2(x, y)) -> D2(z, y)
D2(c1(z), g2(g2(x, y), 0)) -> D2(c1(z), g2(x, y))

The TRS R consists of the following rules:

h2(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, 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(z, g2(x, y)) -> D2(z, y)
D2(c1(z), g2(g2(x, y), 0)) -> D2(c1(z), g2(x, y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 3   
POL(D2(x1, x2)) = 3·x2   
POL(c1(x1)) = 0   
POL(e1(x1)) = 0   
POL(g2(x1, x2)) = 3 + 3·x1 + 3·x2   

The following usable rules [14] were oriented:

g2(e1(x), e1(y)) -> e1(g2(x, y))



↳ 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(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, 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(z, e1(x)) -> H2(c1(z), d2(z, x))

The TRS R consists of the following rules:

h2(z, e1(x)) -> h2(c1(z), d2(z, x))
d2(z, g2(0, 0)) -> e1(0)
d2(z, g2(x, y)) -> g2(e1(x), d2(z, y))
d2(c1(z), g2(g2(x, y), 0)) -> g2(d2(c1(z), g2(x, y)), d2(z, g2(x, y)))
g2(e1(x), e1(y)) -> e1(g2(x, y))

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