Termination w.r.t. Q proof of TRS/AG01#3.8a.trs

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:

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

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

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

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:

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

The set Q consists of the following terms:

pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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)) → QUOT(minus(x, y), s(y))
MINUS(x, s(y)) → PRED(minus(x, y))
LOG(s(s(x))) → QUOT(x, s(s(0)))
LOG(s(s(x))) → LOG(s(quot(x, s(s(0)))))
QUOT(s(x), s(y)) → MINUS(x, y)
MINUS(x, s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
MINUS(x, s(y)) → PRED(minus(x, y))
LOG(s(s(x))) → QUOT(x, s(s(0)))
LOG(s(s(x))) → LOG(s(quot(x, s(s(0)))))
QUOT(s(x), s(y)) → MINUS(x, y)
MINUS(x, s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
MINUS(x, s(y)) → PRED(minus(x, y))
LOG(s(s(x))) → QUOT(x, s(s(0)))
QUOT(s(x), s(y)) → MINUS(x, y)
LOG(s(s(x))) → LOG(s(quot(x, s(s(0)))))
MINUS(x, s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

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

MINUS(x, s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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(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) = MINUS(x1, x2)
s(x1) = s(x1)

Recursive path order with status [2].
Precedence:

s₁ > MINUS₂

Status:

s₁: multiset
MINUS₂: [1,2]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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:

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

The set Q consists of the following terms:

pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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

                  ↳ QDP

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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)) → QUOT(minus(x, y), s(y))

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

Recursive path order with status [2].
Precedence:

QUOT₁ > s₁ > minus₁
0 > minus₁

Status:

minus₁: multiset
QUOT₁: multiset
0: multiset
s₁: multiset

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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:

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

The set Q consists of the following terms:

pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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

                  ↳ QDP

                    ↳ QDPOrderProof

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

LOG(s(s(x))) → LOG(s(quot(x, s(s(0)))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

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.

LOG(s(s(x))) → LOG(s(quot(x, s(s(0)))))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LOG(x1) = x1
s(x1) = s(x1)
quot(x1, x2) = x1
0 = 0
pred(x1) = x1
minus(x1, x2) = x1

Recursive path order with status [2].
Precedence:

trivial

Status:

0: multiset
s₁: multiset

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

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

The set Q consists of the following terms:

pred(s(x0))
minus(x0, 0)
minus(x0, s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(s(0))
log(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.