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(0) -> s1(0)
f1(s1(0)) -> s1(s1(0))
f1(s1(0)) -> *2(s1(s1(0)), f1(0))
f1(+2(x, s1(0))) -> +2(s1(s1(0)), f1(x))
f1(+2(x, y)) -> *2(f1(x), f1(y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(0) -> s1(0)
f1(s1(0)) -> s1(s1(0))
f1(s1(0)) -> *2(s1(s1(0)), f1(0))
f1(+2(x, s1(0))) -> +2(s1(s1(0)), f1(x))
f1(+2(x, y)) -> *2(f1(x), f1(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(s1(0)) -> F1(0)
F1(+2(x, y)) -> F1(y)
F1(+2(x, y)) -> F1(x)
F1(+2(x, s1(0))) -> F1(x)

The TRS R consists of the following rules:

f1(0) -> s1(0)
f1(s1(0)) -> s1(s1(0))
f1(s1(0)) -> *2(s1(s1(0)), f1(0))
f1(+2(x, s1(0))) -> +2(s1(s1(0)), f1(x))
f1(+2(x, y)) -> *2(f1(x), f1(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:

F1(s1(0)) -> F1(0)
F1(+2(x, y)) -> F1(y)
F1(+2(x, y)) -> F1(x)
F1(+2(x, s1(0))) -> F1(x)

The TRS R consists of the following rules:

f1(0) -> s1(0)
f1(s1(0)) -> s1(s1(0))
f1(s1(0)) -> *2(s1(s1(0)), f1(0))
f1(+2(x, s1(0))) -> +2(s1(s1(0)), f1(x))
f1(+2(x, y)) -> *2(f1(x), f1(y))

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

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

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

F1(+2(x, y)) -> F1(y)
F1(+2(x, y)) -> F1(x)
F1(+2(x, s1(0))) -> F1(x)

The TRS R consists of the following rules:

f1(0) -> s1(0)
f1(s1(0)) -> s1(s1(0))
f1(s1(0)) -> *2(s1(s1(0)), f1(0))
f1(+2(x, s1(0))) -> +2(s1(s1(0)), f1(x))
f1(+2(x, y)) -> *2(f1(x), f1(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(+2(x, s1(0))) -> F1(x)
The remaining pairs can at least be oriented weakly.

F1(+2(x, y)) -> F1(y)
F1(+2(x, y)) -> F1(x)
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 2·x1 + 2·x2   
POL(0) = 2   
POL(F1(x1)) = 2·x1   
POL(s1(x1)) = 2·x1   

The following usable rules [14] were oriented: none



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

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

F1(+2(x, y)) -> F1(y)
F1(+2(x, y)) -> F1(x)

The TRS R consists of the following rules:

f1(0) -> s1(0)
f1(s1(0)) -> s1(s1(0))
f1(s1(0)) -> *2(s1(s1(0)), f1(0))
f1(+2(x, s1(0))) -> +2(s1(s1(0)), f1(x))
f1(+2(x, y)) -> *2(f1(x), f1(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(+2(x, y)) -> F1(y)
F1(+2(x, y)) -> F1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(F1(x1)) = 2·x1   

The following usable rules [14] were oriented: none



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

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

f1(0) -> s1(0)
f1(s1(0)) -> s1(s1(0))
f1(s1(0)) -> *2(s1(s1(0)), f1(0))
f1(+2(x, s1(0))) -> +2(s1(s1(0)), f1(x))
f1(+2(x, y)) -> *2(f1(x), f1(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.