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, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

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

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

Q is empty.

The TRS is overlay and locally confluent. By [19] 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, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)


Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

QUOT(x, y) → MINUS(x, y)
QUOT(x, y) → LE(y, x)
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(x, y) → IF_QUOT(minus(x, y), y, le(y, 0), le(y, x))
LE(s(x), s(y)) → LE(x, y)
IF_QUOT(x, y, false, true) → QUOT(x, y)
QUOT(x, y) → LE(y, 0)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

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

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

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

QUOT(x, y) → MINUS(x, y)
QUOT(x, y) → LE(y, x)
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(x, y) → IF_QUOT(minus(x, y), y, le(y, 0), le(y, x))
LE(s(x), s(y)) → LE(x, y)
IF_QUOT(x, y, false, true) → QUOT(x, y)
QUOT(x, y) → LE(y, 0)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 3 less nodes.

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

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP
              ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

R is empty.
The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP
              ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ UsableRulesProof
              ↳ 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, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof
              ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

R is empty.
The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:



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

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

QUOT(x, y) → IF_QUOT(minus(x, y), y, le(y, 0), le(y, x))
IF_QUOT(x, y, false, true) → QUOT(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, y) → if_quot(minus(x, y), y, le(y, 0), le(y, x))
if_quot(x, y, true, z) → divByZeroError
if_quot(x, y, false, true) → s(quot(x, y))
if_quot(x, y, false, false) → 0

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
QDP
                    ↳ QReductionProof

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

QUOT(x, y) → IF_QUOT(minus(x, y), y, le(y, 0), le(y, x))
IF_QUOT(x, y, false, true) → QUOT(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

quot(x0, x1)
if_quot(x0, x1, true, x2)
if_quot(x0, x1, false, true)
if_quot(x0, x1, false, false)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
QDP
                        ↳ Narrowing

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

QUOT(x, y) → IF_QUOT(minus(x, y), y, le(y, 0), le(y, x))
IF_QUOT(x, y, false, true) → QUOT(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule QUOT(x, y) → IF_QUOT(minus(x, y), y, le(y, 0), le(y, x)) at position [2] we obtained the following new rules:

QUOT(y0, s(x0)) → IF_QUOT(minus(y0, s(x0)), s(x0), false, le(s(x0), y0))
QUOT(y0, 0) → IF_QUOT(minus(y0, 0), 0, true, le(0, y0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
QDP
                            ↳ DependencyGraphProof

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

QUOT(y0, 0) → IF_QUOT(minus(y0, 0), 0, true, le(0, y0))
QUOT(y0, s(x0)) → IF_QUOT(minus(y0, s(x0)), s(x0), false, le(s(x0), y0))
IF_QUOT(x, y, false, true) → QUOT(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
QDP
                                ↳ Narrowing

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

QUOT(y0, s(x0)) → IF_QUOT(minus(y0, s(x0)), s(x0), false, le(s(x0), y0))
IF_QUOT(x, y, false, true) → QUOT(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule QUOT(y0, s(x0)) → IF_QUOT(minus(y0, s(x0)), s(x0), false, le(s(x0), y0)) at position [3] we obtained the following new rules:

QUOT(0, s(x0)) → IF_QUOT(minus(0, s(x0)), s(x0), false, false)
QUOT(s(x1), s(x0)) → IF_QUOT(minus(s(x1), s(x0)), s(x0), false, le(x0, x1))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
QDP
                                    ↳ DependencyGraphProof

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

QUOT(0, s(x0)) → IF_QUOT(minus(0, s(x0)), s(x0), false, false)
QUOT(s(x1), s(x0)) → IF_QUOT(minus(s(x1), s(x0)), s(x0), false, le(x0, x1))
IF_QUOT(x, y, false, true) → QUOT(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
QDP
                                        ↳ Narrowing

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

QUOT(s(x1), s(x0)) → IF_QUOT(minus(s(x1), s(x0)), s(x0), false, le(x0, x1))
IF_QUOT(x, y, false, true) → QUOT(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule QUOT(s(x1), s(x0)) → IF_QUOT(minus(s(x1), s(x0)), s(x0), false, le(x0, x1)) at position [0] we obtained the following new rules:

QUOT(s(x0), s(x1)) → IF_QUOT(minus(x0, x1), s(x1), false, le(x1, x0))
QUOT(s(y0), s(y0)) → IF_QUOT(0, s(y0), false, le(y0, y0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Narrowing
QDP
                                            ↳ Instantiation

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

QUOT(s(x0), s(x1)) → IF_QUOT(minus(x0, x1), s(x1), false, le(x1, x0))
QUOT(s(y0), s(y0)) → IF_QUOT(0, s(y0), false, le(y0, y0))
IF_QUOT(x, y, false, true) → QUOT(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF_QUOT(x, y, false, true) → QUOT(x, y) we obtained the following new rules:

IF_QUOT(0, s(z0), false, true) → QUOT(0, s(z0))
IF_QUOT(y_0, s(z1), false, true) → QUOT(y_0, s(z1))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Narrowing
                                          ↳ QDP
                                            ↳ Instantiation
QDP
                                                ↳ DependencyGraphProof

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

IF_QUOT(0, s(z0), false, true) → QUOT(0, s(z0))
IF_QUOT(y_0, s(z1), false, true) → QUOT(y_0, s(z1))
QUOT(s(x0), s(x1)) → IF_QUOT(minus(x0, x1), s(x1), false, le(x1, x0))
QUOT(s(y0), s(y0)) → IF_QUOT(0, s(y0), false, le(y0, y0))

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Narrowing
                                          ↳ QDP
                                            ↳ Instantiation
                                              ↳ QDP
                                                ↳ DependencyGraphProof
QDP
                                                    ↳ ForwardInstantiation

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

IF_QUOT(y_0, s(z1), false, true) → QUOT(y_0, s(z1))
QUOT(s(x0), s(x1)) → IF_QUOT(minus(x0, x1), s(x1), false, le(x1, x0))
QUOT(s(y0), s(y0)) → IF_QUOT(0, s(y0), false, le(y0, y0))

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule IF_QUOT(y_0, s(z1), false, true) → QUOT(y_0, s(z1)) we obtained the following new rules:

IF_QUOT(s(y_0), s(x1), false, true) → QUOT(s(y_0), s(x1))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Narrowing
                                          ↳ QDP
                                            ↳ Instantiation
                                              ↳ QDP
                                                ↳ DependencyGraphProof
                                                  ↳ QDP
                                                    ↳ ForwardInstantiation
QDP
                                                        ↳ DependencyGraphProof

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

QUOT(s(x0), s(x1)) → IF_QUOT(minus(x0, x1), s(x1), false, le(x1, x0))
IF_QUOT(s(y_0), s(x1), false, true) → QUOT(s(y_0), s(x1))
QUOT(s(y0), s(y0)) → IF_QUOT(0, s(y0), false, le(y0, y0))

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Narrowing
                                          ↳ QDP
                                            ↳ Instantiation
                                              ↳ QDP
                                                ↳ DependencyGraphProof
                                                  ↳ QDP
                                                    ↳ ForwardInstantiation
                                                      ↳ QDP
                                                        ↳ DependencyGraphProof
QDP
                                                            ↳ QDPOrderProof

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

QUOT(s(x0), s(x1)) → IF_QUOT(minus(x0, x1), s(x1), false, le(x1, x0))
IF_QUOT(s(y_0), s(x1), false, true) → QUOT(s(y_0), s(x1))

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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.


QUOT(s(x0), s(x1)) → IF_QUOT(minus(x0, x1), s(x1), false, le(x1, x0))
The remaining pairs can at least be oriented weakly.

IF_QUOT(s(y_0), s(x1), false, true) → QUOT(s(y_0), s(x1))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( minus(x1, x2) ) =
/0\
\0/
+
/11\
\01/
·x1+
/00\
\00/
·x2

M( le(x1, x2) ) =
/0\
\0/
+
/00\
\00/
·x1+
/00\
\00/
·x2

M( true ) =
/0\
\0/

M( false ) =
/0\
\0/

M( s(x1) ) =
/0\
\1/
+
/00\
\11/
·x1

M( 0 ) =
/0\
\0/

Tuple symbols:
M( QUOT(x1, x2) ) = 0+
[0,1]
·x1+
[0,0]
·x2

M( IF_QUOT(x1, ..., x4) ) = 0+
[0,1]
·x1+
[0,0]
·x2+
[0,0]
·x3+
[0,0]
·x4


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ Narrowing
                                  ↳ QDP
                                    ↳ DependencyGraphProof
                                      ↳ QDP
                                        ↳ Narrowing
                                          ↳ QDP
                                            ↳ Instantiation
                                              ↳ QDP
                                                ↳ DependencyGraphProof
                                                  ↳ QDP
                                                    ↳ ForwardInstantiation
                                                      ↳ QDP
                                                        ↳ DependencyGraphProof
                                                          ↳ QDP
                                                            ↳ QDPOrderProof
QDP
                                                                ↳ DependencyGraphProof

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

IF_QUOT(s(y_0), s(x1), false, true) → QUOT(s(y_0), s(x1))

The TRS R consists of the following rules:

minus(x, x) → 0
minus(0, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

The set Q consists of the following terms:

minus(x0, x0)
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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.