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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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


F1(c2(s1(x), s1(y))) -> G1(c2(x, y))
G1(c2(s1(x), s1(y))) -> F1(c2(x, y))
The remaining pairs can at least be oriented weakly.

F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))
G1(c2(x, s1(y))) -> G1(c2(s1(x), y))
Used ordering: Polynomial interpretation [21]:

POL(F1(x1)) = x1   
POL(G1(x1)) = 1 + x1   
POL(c2(x1, x2)) = x1 + x2   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

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:

f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(c2(s1(x), s1(y))) -> g1(c2(x, y))
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
g1(c2(s1(x), s1(y))) -> f1(c2(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 2 SCCs.

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

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:

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


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

POL(G1(x1)) = x1   
POL(c2(x1, x2)) = x2   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

f1(c2(s1(x), y)) -> f1(c2(x, s1(y)))
f1(c2(s1(x), s1(y))) -> g1(c2(x, y))
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))
g1(c2(s1(x), s1(y))) -> f1(c2(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
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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:

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


F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F1(x1)) = x1   
POL(c2(x1, x2)) = x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



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

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

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