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(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → Y

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → 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(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → Y

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), s(x1))
geq(x0, 0)
geq(0, s(x0))
geq(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, 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:

DIV(s(X), s(Y)) → IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
DIV(s(X), s(Y)) → MINUS(X, Y)
DIV(s(X), s(Y)) → DIV(minus(X, Y), s(Y))
DIV(s(X), s(Y)) → GEQ(X, Y)
GEQ(s(X), s(Y)) → GEQ(X, Y)
MINUS(s(X), s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → Y

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), s(x1))
geq(x0, 0)
geq(0, s(x0))
geq(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, 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:

DIV(s(X), s(Y)) → IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
DIV(s(X), s(Y)) → MINUS(X, Y)
DIV(s(X), s(Y)) → DIV(minus(X, Y), s(Y))
DIV(s(X), s(Y)) → GEQ(X, Y)
GEQ(s(X), s(Y)) → GEQ(X, Y)
MINUS(s(X), s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → Y

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), s(x1))
geq(x0, 0)
geq(0, s(x0))
geq(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, 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:

DIV(s(X), s(Y)) → IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
DIV(s(X), s(Y)) → DIV(minus(X, Y), s(Y))
DIV(s(X), s(Y)) → MINUS(X, Y)
DIV(s(X), s(Y)) → GEQ(X, Y)
GEQ(s(X), s(Y)) → GEQ(X, Y)
MINUS(s(X), s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → Y

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), s(x1))
geq(x0, 0)
geq(0, s(x0))
geq(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 4 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:

GEQ(s(X), s(Y)) → GEQ(X, Y)

The TRS R consists of the following rules:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → Y

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), s(x1))
geq(x0, 0)
geq(0, s(x0))
geq(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, 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.


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

Recursive path order with status [2].
Quasi-Precedence:
[GEQ1, s1]

Status:
s1: multiset
GEQ1: multiset


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(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → Y

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), s(x1))
geq(x0, 0)
geq(0, s(x0))
geq(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, 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:

MINUS(s(X), s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → Y

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), s(x1))
geq(x0, 0)
geq(0, s(x0))
geq(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, 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(x1)
s(x1)  =  s(x1)

Recursive path order with status [2].
Quasi-Precedence:
[MINUS1, s1]

Status:
MINUS1: multiset
s1: multiset


The following usable rules [14] were oriented: none



↳ 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(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → Y

The set Q consists of the following terms:

minus(0, x0)
minus(s(x0), s(x1))
geq(x0, 0)
geq(0, s(x0))
geq(s(x0), s(x1))
div(0, s(x0))
div(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)

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