Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/AProVESLASHIJCAR

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:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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.

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

DIV₂(div₂(x, y), z) -> TIMES₂(y, z)
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(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)
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)
DIVIDES₂(y, x) -> DIV₂(x, y)
PRIME₁(s₁(s₁(x))) -> PR₂(s₁(s₁(x)), s₁(x))

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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)
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(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)
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)
DIVIDES₂(y, x) -> DIV₂(x, y)
PRIME₁(s₁(s₁(x))) -> PR₂(s₁(s₁(x)), s₁(x))

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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 contains 5 SCCs with 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ 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:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)

Used argument filtering: EQ₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

Used argument filtering: PLUS₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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

            ↳ QDPAfsSolverProof

          ↳ 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:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TIMES₂(s₁(x), y) -> TIMES₂(x, y)

Used argument filtering: TIMES₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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

            ↳ QDPAfsSolverProof

          ↳ 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:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: QUOT₃(x1, x2, x3) = x1
DIV₂(x1, x2) = x1
s₁(x1) = s₁(x1)
div₂(x1, x2) = x1
times₂(x1, x2) = times₂(x1, x2)
0 = 0
plus₂(x1, x2) = plus₂(x1, x2)
Used ordering: Quasi Precedence: times_2 > plus_2 > s_1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

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

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(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:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: QUOT₃(x1, x2, x3) = x1
DIV₂(x1, x2) = x1
div₂(x1, x2) = div₁(x1)
times₂(x1, x2) = times₂(x1, x2)
0 = 0
s₁(x1) = s₁(x1)
plus₂(x1, x2) = plus₂(x1, x2)
Used ordering: Quasi Precedence: times_2 > plus_2 > s_1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ 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:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))

Used argument filtering: QUOT₃(x1, x2, x3) = x2
0 = 0
DIV₂(x1, x2) = x2
s₁(x1) = s
Used ordering: Quasi Precedence: 0 > s

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ 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:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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 contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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:

plus₂(x, 0) -> x
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, 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.