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, 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
  ↳ 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

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:

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

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:

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

Q is empty.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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

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.


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

POL(s(x1)) = 1/2 + (13/4)x_1   
POL(LE(x1, x2)) = (15/4)x_2   
The value of delta used in the strict ordering is 15/8.
The following usable rules [17] were oriented: none



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

Q DP problem:
P is empty.
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.
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
          ↳ 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

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)) = (15/4)x_2   
POL(s(x1)) = 1/2 + (13/4)x_1   
The value of delta used in the strict ordering is 15/8.
The following usable rules [17] were oriented: none



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

Q DP problem:
P is empty.
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.
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
          ↳ QDP

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

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