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:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

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:

-1(s(x), s(y)) → -1(x, y)
F(s(x)) → F(x)
F(s(x)) → G(f(x))
G(s(x)) → G(x)
G(s(x)) → F(g(x))
G(s(x)) → -1(s(x), f(g(x)))
F(s(x)) → -1(s(x), g(f(x)))

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

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:

-1(s(x), s(y)) → -1(x, y)
F(s(x)) → F(x)
F(s(x)) → G(f(x))
G(s(x)) → G(x)
G(s(x)) → F(g(x))
G(s(x)) → -1(s(x), f(g(x)))
F(s(x)) → -1(s(x), g(f(x)))

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

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(s(x), s(y)) → -1(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

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

POL(-1(x1, x2)) = x_1 + (13/4)x_2   
POL(s(x1)) = 5/4 + (15/4)x_1   
The value of delta used in the strict ordering is 85/16.
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
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

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)
F(s(x)) → G(f(x))
G(s(x)) → G(x)
G(s(x)) → F(g(x))

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

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

POL(f(x1)) = (2)x_1   
POL(-(x1, x2)) = x_1   
POL(g(x1)) = 1/4 + (2)x_1   
POL(s(x1)) = 1/4 + (2)x_1   
POL(G(x1)) = 1/2 + (4)x_1   
POL(0) = 1/4   
POL(F(x1)) = 1/4 + (4)x_1   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(x)))

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.