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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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

+1(X, s(Y)) → +1(X, Y)
F(0, s(0), X) → DOUBLE(X)
F(0, s(0), X) → F(X, double(X), X)
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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

+1(X, s(Y)) → +1(X, Y)
F(0, s(0), X) → DOUBLE(X)
F(0, s(0), X) → F(X, double(X), X)
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 [13,14,18] contains 2 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
+1(x1, x2)  =  +1(x1, x2)
s(x1)  =  s(x1)

Recursive path order with status [2].
Quasi-Precedence:
[+^12, s1]

Status:
s1: [1]
+^12: multiset


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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
      ↳ EdgeDeletionProof
        ↳ 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.