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

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:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(x, y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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:

DIV₂(div₂(x, y), z) -> TIMES₂(y, z)
PLUS₂(s₁(x), y) -> PLUS₂(p₁(s₁(x)), y)
PLUS₂(s₁(x), y) -> P₁(s₁(x))
QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
PR₂(x, s₁(s₁(y))) -> DIVIDES₂(s₁(s₁(y)), x)
DIVIDES₂(y, x) -> TIMES₂(div₂(x, y), y)
DIVIDES₂(y, x) -> EQ₂(x, times₂(div₂(x, y), y))
PLUS₂(x, s₁(y)) -> PLUS₂(x, p₁(s₁(y)))
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
TIMES₂(s₁(x), y) -> PLUS₂(y, times₂(x, y))
QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)
PLUS₂(x, s₁(y)) -> P₁(s₁(y))
DIV₂(div₂(x, y), z) -> DIV₂(x, times₂(y, z))
TIMES₂(s₁(x), y) -> TIMES₂(x, y)
PR₂(x, s₁(s₁(y))) -> IF₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
DIV₂(x, y) -> QUOT₃(x, y, y)
IF₃(false, x, y) -> PR₂(x, y)
PRIME₁(s₁(s₁(x))) -> PR₂(s₁(s₁(x)), s₁(x))
DIVIDES₂(y, x) -> DIV₂(x, y)

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(x, 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:

DIV₂(div₂(x, y), z) -> TIMES₂(y, z)
PLUS₂(s₁(x), y) -> PLUS₂(p₁(s₁(x)), y)
PLUS₂(s₁(x), y) -> P₁(s₁(x))
QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
PR₂(x, s₁(s₁(y))) -> DIVIDES₂(s₁(s₁(y)), x)
DIVIDES₂(y, x) -> TIMES₂(div₂(x, y), y)
DIVIDES₂(y, x) -> EQ₂(x, times₂(div₂(x, y), y))
PLUS₂(x, s₁(y)) -> PLUS₂(x, p₁(s₁(y)))
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
TIMES₂(s₁(x), y) -> PLUS₂(y, times₂(x, y))
QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)
PLUS₂(x, s₁(y)) -> P₁(s₁(y))
DIV₂(div₂(x, y), z) -> DIV₂(x, times₂(y, z))
TIMES₂(s₁(x), y) -> TIMES₂(x, y)
PR₂(x, s₁(s₁(y))) -> IF₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
DIV₂(x, y) -> QUOT₃(x, y, y)
IF₃(false, x, y) -> PR₂(x, y)
PRIME₁(s₁(s₁(x))) -> PR₂(s₁(s₁(x)), s₁(x))
DIVIDES₂(y, x) -> DIV₂(x, y)

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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.

EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = x₁ + 1

POL( EQ₂(x₁, x₂) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS₂(s₁(x), y) -> PLUS₂(x, y)
PLUS₂(s₁(x), y) -> PLUS₂(p₁(s₁(x)), y)
PLUS₂(x, s₁(y)) -> PLUS₂(x, p₁(s₁(y)))

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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.

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

The remaining pairs can at least be oriented weakly.

PLUS₂(s₁(x), y) -> PLUS₂(p₁(s₁(x)), y)
PLUS₂(x, s₁(y)) -> PLUS₂(x, p₁(s₁(y)))

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = x₁ + 1

POL( p₁(x₁) ) = x₁

POL( PLUS₂(x₁, x₂) ) = x₁ + x₂ + 1

The following usable rules [14] were oriented:

p₁(s₁(x)) -> x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS₂(s₁(x), y) -> PLUS₂(p₁(s₁(x)), y)
PLUS₂(x, s₁(y)) -> PLUS₂(x, p₁(s₁(y)))

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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.

PLUS₂(s₁(x), y) -> PLUS₂(p₁(s₁(x)), y)

The remaining pairs can at least be oriented weakly.

PLUS₂(x, s₁(y)) -> PLUS₂(x, p₁(s₁(y)))

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = x₁ + 1

POL( p₁(x₁) ) = max{0, x₁ - 1}

POL( PLUS₂(x₁, x₂) ) = x₁ + 1

The following usable rules [14] were oriented:

p₁(s₁(x)) -> x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS₂(x, s₁(y)) -> PLUS₂(x, p₁(s₁(y)))

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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.

PLUS₂(x, s₁(y)) -> PLUS₂(x, p₁(s₁(y)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = x₁ + 1

POL( p₁(x₁) ) = max{0, x₁ - 1}

POL( PLUS₂(x₁, x₂) ) = x₁ + x₂

The following usable rules [14] were oriented:

p₁(s₁(x)) -> x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

TIMES₂(s₁(x), y) -> TIMES₂(x, y)

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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.

TIMES₂(s₁(x), y) -> TIMES₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = x₁ + 1

POL( TIMES₂(x₁, x₂) ) = x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
DIV₂(div₂(x, y), z) -> DIV₂(x, times₂(y, z))
DIV₂(x, y) -> QUOT₃(x, y, y)

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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.

QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
DIV₂(div₂(x, y), z) -> DIV₂(x, times₂(y, z))

The remaining pairs can at least be oriented weakly.

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
DIV₂(x, y) -> QUOT₃(x, y, y)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( QUOT₃(x₁, ..., x₃) ) = x₁ + 1

POL( plus₂(x₁, x₂) ) = max{0, x₁ - 1}

POL( 0 ) = 1

POL( s₁(x₁) ) = x₁ + 1

POL( DIV₂(x₁, x₂) ) = x₁ + 1

POL( times₂(x₁, x₂) ) = max{0, -1}

POL( div₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( p₁(x₁) ) = max{0, -1}

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))
DIV₂(x, y) -> QUOT₃(x, y, y)

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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.

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))

The remaining pairs can at least be oriented weakly.

DIV₂(x, y) -> QUOT₃(x, y, y)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( QUOT₃(x₁, ..., x₃) ) = x₁ + x₂

POL( s₁(x₁) ) = max{0, -1}

POL( 0 ) = 1

POL( DIV₂(x₁, x₂) ) = x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

          ↳ QDP

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

DIV₂(x, y) -> QUOT₃(x, y, y)

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

PR₂(x, s₁(s₁(y))) -> IF₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
IF₃(false, x, y) -> PR₂(x, y)

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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.

PR₂(x, s₁(s₁(y))) -> IF₃(divides₂(s₁(s₁(y)), x), x, s₁(y))

The remaining pairs can at least be oriented weakly.

IF₃(false, x, y) -> PR₂(x, y)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( IF₃(x₁, ..., x₃) ) = x₃

POL( eq₂(x₁, x₂) ) = 1

POL( 0 ) = 1

POL( divides₂(x₁, x₂) ) = max{0, -1}

POL( times₂(x₁, x₂) ) = 1

POL( div₂(x₁, x₂) ) = max{0, x₁ - 1}

POL( PR₂(x₁, x₂) ) = x₂

POL( true ) = max{0, -1}

POL( false ) = max{0, -1}

POL( plus₂(x₁, x₂) ) = max{0, -1}

POL( quot₃(x₁, ..., x₃) ) = x₂

POL( s₁(x₁) ) = x₁ + 1

POL( p₁(x₁) ) = 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

IF₃(false, x, y) -> PR₂(x, y)

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(x)) -> x
plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
plus₂(x, s₁(y)) -> s₁(plus₂(x, p₁(s₁(y))))
times₂(0, y) -> 0
times₂(s₁(0), y) -> y
times₂(s₁(x), y) -> plus₂(y, times₂(x, y))
div₂(0, y) -> 0
div₂(x, y) -> quot₃(x, y, y)
quot₃(0, s₁(y), z) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
quot₃(x, 0, s₁(z)) -> s₁(div₂(x, s₁(z)))
div₂(div₂(x, y), z) -> div₂(x, times₂(y, z))
eq₂(0, 0) -> true
eq₂(s₁(x), 0) -> false
eq₂(0, s₁(y)) -> false
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
divides₂(y, x) -> eq₂(x, times₂(div₂(x, y), y))
prime₁(s₁(s₁(x))) -> pr₂(s₁(s₁(x)), s₁(x))
pr₂(x, s₁(0)) -> true
pr₂(x, s₁(s₁(y))) -> if₃(divides₂(s₁(s₁(y)), x), x, s₁(y))
if₃(true, x, y) -> false
if₃(false, x, y) -> pr₂(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.