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

Q is empty.


↳ QTRS
  ↳ AAECC Innermost

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

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

Q is empty.

We have applied [15,7] to switch to innermost. The TRS R 1 is

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

The TRS R 2 is

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

The signature Sigma is {true, false, le}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

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

QUOT(s(x), s(y)) → MINUS(s(x), s(y))
LE(s(x), s(y)) → LE(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

QUOT(s(x), s(y)) → MINUS(s(x), s(y))
LE(s(x), s(y)) → LE(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

QUOT(s(x), s(y)) → MINUS(s(x), s(y))
LE(s(x), s(y)) → LE(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(s(x), s(y)), s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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, 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(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))

The set Q consists of the following terms:

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


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

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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, 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(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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, 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(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(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)  =  x2
s(x1)  =  s(x1)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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, 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(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(s(x), s(y)), s(y)))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(0, s(x0))
quot(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
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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