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:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

H(cons(X, Y)) → H(g(cons(X, Y)))
F(s(X)) → F(X)
H(cons(X, Y)) → G(cons(X, Y))
G(cons(0, Y)) → G(Y)

The TRS R consists of the following rules:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(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:

H(cons(X, Y)) → H(g(cons(X, Y)))
F(s(X)) → F(X)
H(cons(X, Y)) → G(cons(X, Y))
G(cons(0, Y)) → G(Y)

The TRS R consists of the following rules:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

G(cons(0, Y)) → G(Y)

The TRS R consists of the following rules:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


G(cons(0, Y)) → G(Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x1, x2)) = 1 + (4)x_2   
POL(G(x1)) = (4)x_1   
POL(0) = 1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(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:

F(s(X)) → F(X)

The TRS R consists of the following rules:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(X, Y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(s(X)) → F(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 1 + (4)x_1   
POL(F(x1)) = (4)x_1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

f(s(X)) → f(X)
g(cons(0, Y)) → g(Y)
g(cons(s(X), Y)) → s(X)
h(cons(X, Y)) → h(g(cons(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.