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) -> cons2(0, n__f1(n__s1(n__0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
f1(X) -> n__f1(X)
s1(X) -> n__s1(X)
0 -> n__0
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(X) -> X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(0) -> cons2(0, n__f1(n__s1(n__0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
f1(X) -> n__f1(X)
s1(X) -> n__s1(X)
0 -> n__0
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(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:

ACTIVATE1(n__s1(X)) -> S1(activate1(X))
ACTIVATE1(n__0) -> 01
ACTIVATE1(n__f1(X)) -> F1(activate1(X))
F1(s1(0)) -> P1(s1(0))
F1(s1(0)) -> F1(p1(s1(0)))
ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

f1(0) -> cons2(0, n__f1(n__s1(n__0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
f1(X) -> n__f1(X)
s1(X) -> n__s1(X)
0 -> n__0
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(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:

ACTIVATE1(n__s1(X)) -> S1(activate1(X))
ACTIVATE1(n__0) -> 01
ACTIVATE1(n__f1(X)) -> F1(activate1(X))
F1(s1(0)) -> P1(s1(0))
F1(s1(0)) -> F1(p1(s1(0)))
ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

f1(0) -> cons2(0, n__f1(n__s1(n__0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
f1(X) -> n__f1(X)
s1(X) -> n__s1(X)
0 -> n__0
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(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 2 SCCs with 4 less nodes.

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

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

F1(s1(0)) -> F1(p1(s1(0)))

The TRS R consists of the following rules:

f1(0) -> cons2(0, n__f1(n__s1(n__0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
f1(X) -> n__f1(X)
s1(X) -> n__s1(X)
0 -> n__0
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(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(0)) -> F1(p1(s1(0)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( n__s1(x1) ) = 2


POL( s1(x1) ) = x1 + 2


POL( 0 ) = 3


POL( F1(x1) ) = 2x1


POL( n__0 ) = 3


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



The following usable rules [14] were oriented:

s1(X) -> n__s1(X)
p1(s1(0)) -> 0
0 -> n__0



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

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

f1(0) -> cons2(0, n__f1(n__s1(n__0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
f1(X) -> n__f1(X)
s1(X) -> n__s1(X)
0 -> n__0
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(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.

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

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

ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

f1(0) -> cons2(0, n__f1(n__s1(n__0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
f1(X) -> n__f1(X)
s1(X) -> n__s1(X)
0 -> n__0
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(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.


ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( n__f1(x1) ) = 3x1 + 1


POL( n__s1(x1) ) = 2x1 + 2


POL( ACTIVATE1(x1) ) = 2x1



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:

f1(0) -> cons2(0, n__f1(n__s1(n__0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
f1(X) -> n__f1(X)
s1(X) -> n__s1(X)
0 -> n__0
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(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.