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
  ↳ 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))))

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(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))))

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(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))))

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:

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))))

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(s1(x), s1(y)) -> F2(x, s1(c1(s1(y))))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F2(x1, x2)) = x1   
POL(c1(x1)) = 0   
POL(s1(x1)) = 1 + x1   

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:

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

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(x, c1(y)) -> F2(y, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F2(x1, x2)) = x2   
POL(c1(x1)) = 1 + x1   

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:

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