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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(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:

G(s(x)) → MINUS(s(x), f(g(x)))
F(s(x)) → F(x)
MINUS(s(x), s(y)) → MINUS(x, y)
F(s(x)) → G(f(x))
G(s(x)) → G(x)
F(s(x)) → MINUS(s(x), g(f(x)))
G(s(x)) → F(g(x))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(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:

G(s(x)) → MINUS(s(x), f(g(x)))
F(s(x)) → F(x)
MINUS(s(x), s(y)) → MINUS(x, y)
F(s(x)) → G(f(x))
G(s(x)) → G(x)
F(s(x)) → MINUS(s(x), g(f(x)))
G(s(x)) → F(g(x))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(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:

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(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.


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

POL(MINUS(x1, x2)) = (1/2)x_1   
POL(s(x1)) = 1/2 + (4)x_1   
The value of delta used in the strict ordering is 1/4.
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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(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)
The remaining pairs can at least be oriented weakly.

G(s(x)) → F(g(x))
Used ordering: Polynomial interpretation [25,35]:

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

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



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

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

G(s(x)) → F(g(x))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(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 0 SCCs with 1 less node.