Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

a -> g1(c)
g1(a) -> b
f2(g1(X), b) -> f2(a, X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a -> g1(c)
g1(a) -> b
f2(g1(X), b) -> f2(a, 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:

F2(g1(X), b) -> A
F2(g1(X), b) -> F2(a, X)
A -> G1(c)

The TRS R consists of the following rules:

a -> g1(c)
g1(a) -> b
f2(g1(X), b) -> f2(a, 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:

F2(g1(X), b) -> A
F2(g1(X), b) -> F2(a, X)
A -> G1(c)

The TRS R consists of the following rules:

a -> g1(c)
g1(a) -> b
f2(g1(X), b) -> f2(a, 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 2 less nodes.

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

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

F2(g1(X), b) -> F2(a, X)

The TRS R consists of the following rules:

a -> g1(c)
g1(a) -> b
f2(g1(X), b) -> f2(a, 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.


F2(g1(X), b) -> F2(a, X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( b ) = 3


POL( F2(x1, x2) ) = 3x1 + 3x2 + 2


POL( a ) = max{0, -3}


POL( g1(x1) ) = 2x1


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



The following usable rules [14] were oriented:

a -> g1(c)



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

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

a -> g1(c)
g1(a) -> b
f2(g1(X), b) -> f2(a, X)

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.