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:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(X), X)
g(X, Y) → X
g(X, Y) → Y

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(X), X)
g(X, Y) → X
g(X, Y) → 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:

F(0, s(0), X) → F(X, double(X), X)
F(0, s(0), X) → DOUBLE(X)
+1(X, s(Y)) → +1(X, Y)
DOUBLE(X) → +1(X, X)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(X), X)
g(X, Y) → X
g(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:

F(0, s(0), X) → F(X, double(X), X)
F(0, s(0), X) → DOUBLE(X)
+1(X, s(Y)) → +1(X, Y)
DOUBLE(X) → +1(X, X)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(X), X)
g(X, Y) → X
g(X, Y) → 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:

+1(X, s(Y)) → +1(X, Y)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(X), X)
g(X, Y) → X
g(X, Y) → 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.


+1(X, s(Y)) → +1(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 1 + (4)x_1   
POL(+1(x1, x2)) = (4)x_2   
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:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(X), X)
g(X, Y) → X
g(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:

F(0, s(0), X) → F(X, double(X), X)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(X), X)
g(X, Y) → X
g(X, Y) → Y

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