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(s1(x)) -> s1(f1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(s1(x)) -> s1(f1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x

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(x)) -> F1(f1(p1(s1(x))))
F1(s1(x)) -> P1(s1(x))
F1(s1(x)) -> F1(p1(s1(x)))

The TRS R consists of the following rules:

f1(s1(x)) -> s1(f1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x

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(x)) -> F1(f1(p1(s1(x))))
F1(s1(x)) -> P1(s1(x))
F1(s1(x)) -> F1(p1(s1(x)))

The TRS R consists of the following rules:

f1(s1(x)) -> s1(f1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x

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(s1(x)) -> F1(f1(p1(s1(x))))
F1(s1(x)) -> F1(p1(s1(x)))

The TRS R consists of the following rules:

f1(s1(x)) -> s1(f1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x

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(s1(x)) -> F1(f1(p1(s1(x))))
F1(s1(x)) -> F1(p1(s1(x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F1(x1) ) = max{0, x1 - 1}


POL( s1(x1) ) = x1 + 2


POL( f1(x1) ) = x1


POL( p1(x1) ) = max{0, x1 - 2}


POL( 0 ) = 0



The following usable rules [14] were oriented:

f1(0) -> 0
p1(s1(x)) -> x
f1(s1(x)) -> s1(f1(f1(p1(s1(x)))))



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

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

f1(s1(x)) -> s1(f1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x

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.