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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(s1(x), s1(y)), s1(y)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(s1(x), s1(y)), s1(y)))

Q is empty.

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

QUOT2(s1(x), s1(y)) -> MINUS2(s1(x), s1(y))
LE2(s1(x), s1(y)) -> LE2(x, y)
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
QUOT2(s1(x), s1(y)) -> QUOT2(minus2(s1(x), s1(y)), s1(y))

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(s1(x), s1(y)), s1(y)))

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:

QUOT2(s1(x), s1(y)) -> MINUS2(s1(x), s1(y))
LE2(s1(x), s1(y)) -> LE2(x, y)
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
QUOT2(s1(x), s1(y)) -> QUOT2(minus2(s1(x), s1(y)), s1(y))

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(s1(x), s1(y)), s1(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 1 less node.

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

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

LE2(s1(x), s1(y)) -> LE2(x, y)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(s1(x), s1(y)), s1(y)))

Q is empty.
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.


LE2(s1(x), s1(y)) -> LE2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( LE2(x1, x2) ) = x2


POL( s1(x1) ) = x1 + 1



The following usable rules [14] 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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(s1(x), s1(y)), s1(y)))

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:

MINUS2(s1(x), s1(y)) -> MINUS2(x, y)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(s1(x), s1(y)), s1(y)))

Q is empty.
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.


MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS2(x1, x2) ) = x2


POL( s1(x1) ) = x1 + 1



The following usable rules [14] 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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(s1(x), s1(y)), s1(y)))

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:

QUOT2(s1(x), s1(y)) -> QUOT2(minus2(s1(x), s1(y)), s1(y))

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(s1(x), s1(y)), s1(y)))

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