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:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

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:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, 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:

PLUS(s(x), y) → PLUS(x, y)
QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → PLUS(z, s(0))
QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))

The TRS R consists of the following rules:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, s(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:

PLUS(s(x), y) → PLUS(x, y)
QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → PLUS(z, s(0))
QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))

The TRS R consists of the following rules:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, s(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:

PLUS(s(x), y) → PLUS(x, y)
QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → PLUS(z, s(0))
QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))

The TRS R consists of the following rules:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ 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:

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, 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.


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

Recursive Path Order [2].
Precedence:
trivial

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:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, 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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof

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

QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))

The TRS R consists of the following rules:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, 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.


QUOT(s(x), s(y), z) → QUOT(x, y, z)
The remaining pairs can at least be oriented weakly.

QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2, x3)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof

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

QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))

The TRS R consists of the following rules:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, 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.


QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2, x3)  =  x2
0  =  0
plus(x1, x2)  =  x2
s(x1)  =  s

Recursive Path Order [2].
Precedence:
0 > s

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), s(z)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, 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.