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(X) -> cons2(X, n__f1(g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(X) -> X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(X) -> cons2(X, n__f1(g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
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:

SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
F1(X) -> G1(X)
ACTIVATE1(n__f1(X)) -> F1(X)
G1(s1(X)) -> G1(X)

The TRS R consists of the following rules:

f1(X) -> cons2(X, n__f1(g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
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:

SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
F1(X) -> G1(X)
ACTIVATE1(n__f1(X)) -> F1(X)
G1(s1(X)) -> G1(X)

The TRS R consists of the following rules:

f1(X) -> cons2(X, n__f1(g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
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 3 less nodes.

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

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

G1(s1(X)) -> G1(X)

The TRS R consists of the following rules:

f1(X) -> cons2(X, n__f1(g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
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.


G1(s1(X)) -> G1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = 2x1 + 1


POL( G1(x1) ) = x1 + 2



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:

f1(X) -> cons2(X, n__f1(g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
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:

SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))

The TRS R consists of the following rules:

f1(X) -> cons2(X, n__f1(g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
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.


SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( SEL2(x1, x2) ) = max{0, x1 - 1}


POL( f1(x1) ) = x1 + 1


POL( 0 ) = 2


POL( s1(x1) ) = 2x1 + 2


POL( n__f1(x1) ) = max{0, 2x1 - 2}


POL( g1(x1) ) = max{0, -1}


POL( cons2(x1, x2) ) = 2x1 + 2


POL( activate1(x1) ) = max{0, -2}



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(X) -> cons2(X, n__f1(g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
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.