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:

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> 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:

F1(f1(x)) -> F1(d1(f1(x)))
F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))
G1(c2(x, s1(y))) -> G1(c2(s1(x), y))

The TRS R consists of the following rules:

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> 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:

F1(f1(x)) -> F1(d1(f1(x)))
F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))
G1(c2(x, s1(y))) -> G1(c2(s1(x), y))

The TRS R consists of the following rules:

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

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

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

F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))

The TRS R consists of the following rules:

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> 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(c2(s1(x), y)) -> F1(c2(x, s1(y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F1(x1) ) = max{0, 2x1 - 1}


POL( s1(x1) ) = 3x1 + 1


POL( c2(x1, x2) ) = max{0, 2x1 - 1}



The following usable rules [14] were oriented: none



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

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

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> 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.

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

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

G1(c2(x, s1(y))) -> G1(c2(s1(x), y))

The TRS R consists of the following rules:

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> 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.


G1(c2(x, s1(y))) -> G1(c2(s1(x), y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 2


POL( c2(x1, x2) ) = 2x2


POL( G1(x1) ) = max{0, x1 - 3}



The following usable rules [14] were oriented: none



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

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

g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(f1(x)) -> f1(d1(f1(x)))
f1(x) -> 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.