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:

f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))

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:

G1(x) -> F1(g1(x))
F1(g1(a)) -> G1(b)
G1(x) -> G1(x)
F1(g1(a)) -> F1(s1(g1(b)))

The TRS R consists of the following rules:

f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))

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:

G1(x) -> F1(g1(x))
F1(g1(a)) -> G1(b)
G1(x) -> G1(x)
F1(g1(a)) -> F1(s1(g1(b)))

The TRS R consists of the following rules:

f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

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

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

G1(x) -> F1(g1(x))
F1(g1(a)) -> G1(b)
G1(x) -> G1(x)

The TRS R consists of the following rules:

f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))

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.


F1(g1(a)) -> G1(b)
The remaining pairs can at least be oriented weakly.

G1(x) -> F1(g1(x))
G1(x) -> G1(x)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( b ) = max{0, -1}


POL( G1(x1) ) = x1 + 1


POL( a ) = 1


POL( f1(x1) ) = max{0, -1}


POL( s1(x1) ) = max{0, -1}


POL( g1(x1) ) = x1


POL( F1(x1) ) = x1 + 1



The following usable rules [14] were oriented:

f1(f1(x)) -> b
g1(x) -> f1(g1(x))
f1(g1(a)) -> f1(s1(g1(b)))



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

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

G1(x) -> F1(g1(x))
G1(x) -> G1(x)

The TRS R consists of the following rules:

f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

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

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

G1(x) -> G1(x)

The TRS R consists of the following rules:

f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))

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