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:

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(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:

P1(s1(s1(x))) -> P1(s1(x))
FAC1(s1(x)) -> P1(s1(x))
FAC1(s1(x)) -> FAC1(p1(s1(x)))

The TRS R consists of the following rules:

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(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:

P1(s1(s1(x))) -> P1(s1(x))
FAC1(s1(x)) -> P1(s1(x))
FAC1(s1(x)) -> FAC1(p1(s1(x)))

The TRS R consists of the following rules:

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))

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:

P1(s1(s1(x))) -> P1(s1(x))

The TRS R consists of the following rules:

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(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.


P1(s1(s1(x))) -> P1(s1(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( P1(x1) ) = max{0, x1 - 1}


POL( s1(x1) ) = x1 + 1



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:

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(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.

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

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

FAC1(s1(x)) -> FAC1(p1(s1(x)))

The TRS R consists of the following rules:

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(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.


FAC1(s1(x)) -> FAC1(p1(s1(x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( FAC1(x1) ) = x1


POL( s1(x1) ) = x1 + 1


POL( p1(x1) ) = max{0, x1 - 1}


POL( 0 ) = 0



The following usable rules [14] were oriented:

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



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

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

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(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.