Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/AProVESLASHIJCAR

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:

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.

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)
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 [13,14,18] contains 5 SCCs with 7 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:

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
EQ₂(x1, x2) = EQ₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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:

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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PLUS₂(x1, x2) = PLUS₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

[PLUS₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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

            ↳ 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:

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

TIMES₂(s₁(x), y) -> TIMES₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
TIMES₂(x1, x2) = TIMES₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

[TIMES₁, s_1]

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:

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

            ↳ 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:

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.

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

Used ordering: Combined order from the following AFS and order.
QUOT₃(x1, x2, x3) = x1
0 = 0
s₁(x1) = x1
DIV₂(x1, x2) = x1
div₂(x1, x2) = div₂(x1, x2)
times₂(x1, x2) = times
plus₂(x1, x2) = plus

Lexicographic Path Order [19].
Precedence:

trivial

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(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.

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

Used ordering: Combined order from the following AFS and order.
QUOT₃(x1, x2, x3) = QUOT₂(x1, x3)
s₁(x1) = s₁(x1)
0 = 0
DIV₂(x1, x2) = DIV₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

[QUOT₂, s₁, 0, DIV_2]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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:

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

QUOT₃(x, 0, s₁(z)) -> DIV₂(x, s₁(z))

The remaining pairs can at least by weakly be oriented.

DIV₂(x, y) -> QUOT₃(x, y, y)

Used ordering: Combined order from the following AFS and order.
QUOT₃(x1, x2, x3) = QUOT₂(x1, x2)
0 = 0
s₁(x1) = s
DIV₂(x1, x2) = DIV₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

[QUOT₂, 0, DIV_2] > s

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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:

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 [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:

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.

IF₃(false, x, y) -> PR₂(x, y)

Used ordering: Combined order from the following AFS and order.
PR₂(x1, x2) = x2
s₁(x1) = s₁(x1)
IF₃(x1, x2, x3) = x3
divides₂(x1, x2) = divides
false = false
times₂(x1, x2) = times₁(x2)
0 = 0
plus₂(x1, x2) = plus₁(x2)
quot₃(x1, x2, x3) = quot₁(x1)
div₂(x1, x2) = div₂(x1, x2)
eq₂(x1, x2) = x2
true = true

Lexicographic Path Order [19].
Precedence:

divides > [s₁, times₁, div_2]
0 > false > [s₁, times₁, div_2]
0 > true > [s₁, times₁, div_2]
plus₁ > [s₁, times₁, div_2]
quot₁ > [s₁, times₁, div_2]

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:

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 [13,14,18] contains 0 SCCs with 1 less node.