Termination w.r.t. Q proof of TRS/AProVEIJCAR

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:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, 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:

DIV(div(x, y), z) → TIMES(y, z)
PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(s(x), y) → P(s(x))
QUOT(x, 0, s(z)) → DIV(x, s(z))
PR(x, s(s(y))) → DIVIDES(s(s(y)), x)
DIVIDES(y, x) → TIMES(div(x, y), y)
DIVIDES(y, x) → EQ(x, times(div(x, y), y))
PLUS(x, s(y)) → PLUS(x, p(s(y)))
PLUS(s(x), y) → PLUS(x, y)
TIMES(s(x), y) → PLUS(y, times(x, y))
QUOT(s(x), s(y), z) → QUOT(x, y, z)
EQ(s(x), s(y)) → EQ(x, y)
PLUS(x, s(y)) → P(s(y))
DIV(div(x, y), z) → DIV(x, times(y, z))
TIMES(s(x), y) → TIMES(x, y)
PR(x, s(s(y))) → IF(divides(s(s(y)), x), x, s(y))
DIV(x, y) → QUOT(x, y, y)
IF(false, x, y) → PR(x, y)
PRIME(s(s(x))) → PR(s(s(x)), s(x))
DIVIDES(y, x) → DIV(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

DIV(div(x, y), z) → TIMES(y, z)
PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(s(x), y) → P(s(x))
QUOT(x, 0, s(z)) → DIV(x, s(z))
PR(x, s(s(y))) → DIVIDES(s(s(y)), x)
DIVIDES(y, x) → TIMES(div(x, y), y)
DIVIDES(y, x) → EQ(x, times(div(x, y), y))
PLUS(x, s(y)) → PLUS(x, p(s(y)))
PLUS(s(x), y) → PLUS(x, y)
TIMES(s(x), y) → PLUS(y, times(x, y))
QUOT(s(x), s(y), z) → QUOT(x, y, z)
EQ(s(x), s(y)) → EQ(x, y)
PLUS(x, s(y)) → P(s(y))
DIV(div(x, y), z) → DIV(x, times(y, z))
TIMES(s(x), y) → TIMES(x, y)
PR(x, s(s(y))) → IF(divides(s(s(y)), x), x, s(y))
DIV(x, y) → QUOT(x, y, y)
IF(false, x, y) → PR(x, y)
PRIME(s(s(x))) → PR(s(s(x)), s(x))
DIVIDES(y, x) → DIV(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

DIV(div(x, y), z) → TIMES(y, z)
PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(s(x), y) → P(s(x))
QUOT(x, 0, s(z)) → DIV(x, s(z))
PR(x, s(s(y))) → DIVIDES(s(s(y)), x)
DIVIDES(y, x) → TIMES(div(x, y), y)
PLUS(x, s(y)) → PLUS(x, p(s(y)))
DIVIDES(y, x) → EQ(x, times(div(x, y), y))
PLUS(s(x), y) → PLUS(x, y)
TIMES(s(x), y) → PLUS(y, times(x, y))
EQ(s(x), s(y)) → EQ(x, y)
QUOT(s(x), s(y), z) → QUOT(x, y, z)
PLUS(x, s(y)) → P(s(y))
DIV(div(x, y), z) → DIV(x, times(y, z))
TIMES(s(x), y) → TIMES(x, y)
PR(x, s(s(y))) → IF(divides(s(s(y)), x), x, s(y))
DIV(x, y) → QUOT(x, y, y)
DIVIDES(y, x) → DIV(x, y)
PRIME(s(s(x))) → PR(s(s(x)), s(x))
IF(false, x, y) → PR(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, 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.

EQ(s(x), s(y)) → EQ(x, y)

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

Recursive path order with status [2].
Precedence:

s₁ > EQ₁

Status:

EQ₁: multiset
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, 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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(x, s(y)) → PLUS(x, p(s(y)))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, 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.

PLUS(s(x), y) → PLUS(x, y)

The remaining pairs can at least be oriented weakly.

PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(x, s(y)) → PLUS(x, p(s(y)))

Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2) = PLUS(x1)
s(x1) = s(x1)
p(x1) = x1

Recursive path order with status [2].
Precedence:

trivial

Status:

s₁: multiset
PLUS₁: [1]

The following usable rules [14] were oriented:

p(s(x)) → x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PLUS(s(x), y) → PLUS(p(s(x)), y)
PLUS(x, s(y)) → PLUS(x, p(s(y)))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, 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.

TIMES(s(x), y) → TIMES(x, y)

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

Recursive path order with status [2].
Precedence:

s₁ > TIMES₁

Status:

TIMES₁: multiset
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, 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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

QUOT(x, 0, s(z)) → DIV(x, s(z))
QUOT(s(x), s(y), z) → QUOT(x, y, z)
DIV(div(x, y), z) → DIV(x, times(y, z))
DIV(x, y) → QUOT(x, y, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, 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.

QUOT(s(x), s(y), z) → QUOT(x, y, z)
DIV(div(x, y), z) → DIV(x, times(y, z))

The remaining pairs can at least be oriented weakly.

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

Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2, x3) = x1
0 = 0
s(x1) = s(x1)
DIV(x1, x2) = x1
div(x1, x2) = div(x1)
times(x1, x2) = times(x1)
plus(x1, x2) = plus(x1)
p(x1) = x1

Recursive path order with status [2].
Precedence:

s₁ > plus₁
div₁ > times₁ > 0
div₁ > times₁ > plus₁

Status:

plus₁: multiset
0: multiset
s₁: multiset
times₁: multiset
div₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

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

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

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, 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.

QUOT(x, 0, s(z)) → DIV(x, s(z))

The remaining pairs can at least be oriented weakly.

DIV(x, y) → QUOT(x, y, y)

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

Recursive path order with status [2].
Precedence:

0 > s

Status:

0: multiset
s: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

              ↳ QDP

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

DIV(x, y) → QUOT(x, y, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

PR(x, s(s(y))) → IF(divides(s(s(y)), x), x, s(y))
IF(false, x, y) → PR(x, y)

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, 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.

PR(x, s(s(y))) → IF(divides(s(s(y)), x), x, s(y))
IF(false, x, y) → PR(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PR(x1, x2) = PR(x2)
s(x1) = s(x1)
IF(x1, x2, x3) = IF(x3)
divides(x1, x2) = x1
false = false
p(x1) = p(x1)
eq(x1, x2) = eq
0 = 0
plus(x1, x2) = plus(x1)
quot(x1, x2, x3) = quot(x1, x2)
times(x1, x2) = times(x1)
div(x1, x2) = div(x1, x2)
true = true

Recursive path order with status [2].
Precedence:

s₁ > IF₁ > PR₁ > p₁
s₁ > eq > true > p₁
s₁ > 0 > div₂ > p₁
s₁ > 0 > true > p₁
false > PR₁ > p₁
plus₁ > p₁
quot₂ > 0 > div₂ > p₁
quot₂ > 0 > true > p₁
times₁ > 0 > div₂ > p₁
times₁ > 0 > true > p₁

Status:

eq: []
plus₁: multiset
quot₂: multiset
IF₁: multiset
0: multiset
div₂: multiset
true: multiset
false: multiset
s₁: multiset
times₁: multiset
PR₁: multiset
p₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

p(0) → 0
p(s(x)) → x
plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → s(plus(p(s(x)), y))
plus(x, s(y)) → s(plus(x, p(s(y))))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(y)) → false
eq(s(x), s(y)) → eq(x, y)
divides(y, x) → eq(x, times(div(x, y), y))
prime(s(s(x))) → pr(s(s(x)), s(x))
pr(x, s(0)) → true
pr(x, s(s(y))) → if(divides(s(s(y)), x), x, s(y))
if(true, x, y) → false
if(false, x, y) → pr(x, 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.