Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

+2(X, 0) -> X
+2(X, s1(Y)) -> s1(+2(X, Y))
double1(X) -> +2(X, X)
f3(0, s1(0), X) -> f3(X, double1(X), X)
g2(X, Y) -> X
g2(X, Y) -> Y

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+2(X, 0) -> X
+2(X, s1(Y)) -> s1(+2(X, Y))
double1(X) -> +2(X, X)
f3(0, s1(0), X) -> f3(X, double1(X), X)
g2(X, Y) -> X
g2(X, Y) -> 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:

+12(X, s1(Y)) -> +12(X, Y)
F3(0, s1(0), X) -> DOUBLE1(X)
F3(0, s1(0), X) -> F3(X, double1(X), X)
DOUBLE1(X) -> +12(X, X)

The TRS R consists of the following rules:

+2(X, 0) -> X
+2(X, s1(Y)) -> s1(+2(X, Y))
double1(X) -> +2(X, X)
f3(0, s1(0), X) -> f3(X, double1(X), X)
g2(X, Y) -> X
g2(X, Y) -> 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:

+12(X, s1(Y)) -> +12(X, Y)
F3(0, s1(0), X) -> DOUBLE1(X)
F3(0, s1(0), X) -> F3(X, double1(X), X)
DOUBLE1(X) -> +12(X, X)

The TRS R consists of the following rules:

+2(X, 0) -> X
+2(X, s1(Y)) -> s1(+2(X, Y))
double1(X) -> +2(X, X)
f3(0, s1(0), X) -> f3(X, double1(X), X)
g2(X, Y) -> X
g2(X, Y) -> 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:

+12(X, s1(Y)) -> +12(X, Y)

The TRS R consists of the following rules:

+2(X, 0) -> X
+2(X, s1(Y)) -> s1(+2(X, Y))
double1(X) -> +2(X, X)
f3(0, s1(0), X) -> f3(X, double1(X), X)
g2(X, Y) -> X
g2(X, Y) -> 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.


+12(X, s1(Y)) -> +12(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( +12(x1, x2) ) = 2x1 + 2x2 + 2


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

+2(X, 0) -> X
+2(X, s1(Y)) -> s1(+2(X, Y))
double1(X) -> +2(X, X)
f3(0, s1(0), X) -> f3(X, double1(X), X)
g2(X, Y) -> X
g2(X, Y) -> 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

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

F3(0, s1(0), X) -> F3(X, double1(X), X)

The TRS R consists of the following rules:

+2(X, 0) -> X
+2(X, s1(Y)) -> s1(+2(X, Y))
double1(X) -> +2(X, X)
f3(0, s1(0), X) -> f3(X, double1(X), X)
g2(X, Y) -> X
g2(X, Y) -> Y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.