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)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, 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:

H1(cons2(X, Y)) -> G1(cons2(X, Y))
G1(cons2(0, Y)) -> G1(Y)
H1(cons2(X, Y)) -> H1(g1(cons2(X, Y)))
F1(s1(X)) -> F1(X)

The TRS R consists of the following rules:

f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, 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:

H1(cons2(X, Y)) -> G1(cons2(X, Y))
G1(cons2(0, Y)) -> G1(Y)
H1(cons2(X, Y)) -> H1(g1(cons2(X, Y)))
F1(s1(X)) -> F1(X)

The TRS R consists of the following rules:

f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, Y)))

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 2 less nodes.

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

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

G1(cons2(0, Y)) -> G1(Y)

The TRS R consists of the following rules:

f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, 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.


G1(cons2(0, Y)) -> G1(Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( cons2(x1, x2) ) = x2 + 1


POL( 0 ) = max{0, -3}


POL( G1(x1) ) = x1



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(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, 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.

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

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

F1(s1(X)) -> F1(X)

The TRS R consists of the following rules:

f1(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, 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(s1(X)) -> F1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = 3x1 + 3


POL( F1(x1) ) = 2x1 + 3



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(s1(X)) -> f1(X)
g1(cons2(0, Y)) -> g1(Y)
g1(cons2(s1(X), Y)) -> s1(X)
h1(cons2(X, Y)) -> h1(g1(cons2(X, 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.