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)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))

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)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))

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)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(s(x0), s(x1))


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

MINUS(s(X), s(Y)) → P(minus(X, Y))
DIV(s(X), s(Y)) → MINUS(X, Y)
DIV(s(X), s(Y)) → DIV(minus(X, Y), s(Y))
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)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(s(x0), s(x1))

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:

MINUS(s(X), s(Y)) → P(minus(X, Y))
DIV(s(X), s(Y)) → MINUS(X, Y)
DIV(s(X), s(Y)) → DIV(minus(X, Y), s(Y))
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)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(s(x0), s(x1))

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:

MINUS(s(X), s(Y)) → P(minus(X, Y))
DIV(s(X), s(Y)) → DIV(minus(X, Y), s(Y))
DIV(s(X), s(Y)) → MINUS(X, Y)
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)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(s(x0), s(x1))

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)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(s(x0), s(x1))

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)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(s(x0), s(x1))

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
                    ↳ QDPOrderProof

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

DIV(s(X), s(Y)) → DIV(minus(X, Y), s(Y))

The TRS R consists of the following rules:

minus(X, 0) → X
minus(s(X), s(Y)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(s(x0), s(x1))

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.


DIV(s(X), s(Y)) → DIV(minus(X, Y), s(Y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
DIV(x1, x2)  =  DIV(x1)
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
p(x1)  =  x1
0  =  0

Recursive Path Order [2].
Precedence:
[DIV1, s1]


The following usable rules [14] were oriented:

p(s(X)) → X
minus(X, 0) → X
minus(s(X), s(Y)) → p(minus(X, Y))



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

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

minus(X, 0) → X
minus(s(X), s(Y)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
p(s(x0))
div(0, s(x0))
div(s(x0), s(x1))

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