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:

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), s1(y)) -> f2(x, s1(c1(s1(y))))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), s1(y)) -> f2(x, s1(c1(s1(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:

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), s1(y)) -> f2(x, s1(c1(s1(y))))

The set Q consists of the following terms:

f2(x0, c1(x1))
f2(s1(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:

F2(s1(x), s1(y)) -> F2(x, s1(c1(s1(y))))
F2(x, c1(y)) -> F2(y, y)
F2(x, c1(y)) -> F2(x, s1(f2(y, y)))

The TRS R consists of the following rules:

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), s1(y)) -> f2(x, s1(c1(s1(y))))

The set Q consists of the following terms:

f2(x0, c1(x1))
f2(s1(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:

F2(s1(x), s1(y)) -> F2(x, s1(c1(s1(y))))
F2(x, c1(y)) -> F2(y, y)
F2(x, c1(y)) -> F2(x, s1(f2(y, y)))

The TRS R consists of the following rules:

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), s1(y)) -> f2(x, s1(c1(s1(y))))

The set Q consists of the following terms:

f2(x0, c1(x1))
f2(s1(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 with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

F2(s1(x), s1(y)) -> F2(x, s1(c1(s1(y))))

The TRS R consists of the following rules:

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), s1(y)) -> f2(x, s1(c1(s1(y))))

The set Q consists of the following terms:

f2(x0, c1(x1))
f2(s1(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.


F2(s1(x), s1(y)) -> F2(x, s1(c1(s1(y))))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F2(x1, x2)  =  F1(x1)
s1(x1)  =  s1(x1)
c1(x1)  =  c1(x1)

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

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:

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), s1(y)) -> f2(x, s1(c1(s1(y))))

The set Q consists of the following terms:

f2(x0, c1(x1))
f2(s1(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:

F2(x, c1(y)) -> F2(y, y)

The TRS R consists of the following rules:

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), s1(y)) -> f2(x, s1(c1(s1(y))))

The set Q consists of the following terms:

f2(x0, c1(x1))
f2(s1(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.


F2(x, c1(y)) -> F2(y, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F2(x1, x2)  =  F2(x1, x2)
c1(x1)  =  c1(x1)

Lexicographic Path Order [19].
Precedence:
c1 > F2

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:

f2(x, c1(y)) -> f2(x, s1(f2(y, y)))
f2(s1(x), s1(y)) -> f2(x, s1(c1(s1(y))))

The set Q consists of the following terms:

f2(x0, c1(x1))
f2(s1(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.