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:

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
  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ 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)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))


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

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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.


MINUS(s(x), s(y)) → MINUS(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x2)
s(x1)  =  s(x1)

Recursive Path Order [2].
Precedence:
s1 > MINUS1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ 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:

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)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP

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

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
f(0)
f(s(x0))
g(0)
g(s(x0))

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