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

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

Q restricted rewrite system:
The TRS R 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))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

f1(c2(s1(x0), x1))
g1(c2(x0, s1(x1)))


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

The set Q consists of the following terms:

f1(c2(s1(x0), x1))
g1(c2(x0, s1(x1)))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ 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)))
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))

The set Q consists of the following terms:

f1(c2(s1(x0), x1))
g1(c2(x0, s1(x1)))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ 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)))
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))

The set Q consists of the following terms:

f1(c2(s1(x0), x1))
g1(c2(x0, s1(x1)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


G1(c2(x, s1(y))) -> G1(c2(s1(x), y))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
G1(x1)  =  G1(x1)
c2(x1, x2)  =  x2
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[G1, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ 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)))
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))

The set Q consists of the following terms:

f1(c2(s1(x0), x1))
g1(c2(x0, s1(x1)))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ 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)))
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))

The set Q consists of the following terms:

f1(c2(s1(x0), x1))
g1(c2(x0, s1(x1)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


F1(c2(s1(x), y)) -> F1(c2(x, s1(y)))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F1(x1)  =  F1(x1)
c2(x1, x2)  =  c1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[F1, c1] > s1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ 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)))
g1(c2(x, s1(y))) -> g1(c2(s1(x), y))

The set Q consists of the following terms:

f1(c2(s1(x0), x1))
g1(c2(x0, s1(x1)))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.