Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/SK90SLASH2.29.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:

prime₁(0) -> false
prime₁(s₁(0)) -> false
prime₁(s₁(s₁(x))) -> prime1₂(s₁(s₁(x)), s₁(x))
prime1₂(x, 0) -> false
prime1₂(x, s₁(0)) -> true
prime1₂(x, s₁(s₁(y))) -> and₂(not₁(divp₂(s₁(s₁(y)), x)), prime1₂(x, s₁(y)))
divp₂(x, y) -> =₂(rem₂(x, y), 0)

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

prime₁(0) -> false
prime₁(s₁(0)) -> false
prime₁(s₁(s₁(x))) -> prime1₂(s₁(s₁(x)), s₁(x))
prime1₂(x, 0) -> false
prime1₂(x, s₁(0)) -> true
prime1₂(x, s₁(s₁(y))) -> and₂(not₁(divp₂(s₁(s₁(y)), x)), prime1₂(x, s₁(y)))
divp₂(x, y) -> =₂(rem₂(x, y), 0)

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

prime₁(0) -> false
prime₁(s₁(0)) -> false
prime₁(s₁(s₁(x))) -> prime1₂(s₁(s₁(x)), s₁(x))
prime1₂(x, 0) -> false
prime1₂(x, s₁(0)) -> true
prime1₂(x, s₁(s₁(y))) -> and₂(not₁(divp₂(s₁(s₁(y)), x)), prime1₂(x, s₁(y)))
divp₂(x, y) -> =₂(rem₂(x, y), 0)

The set Q consists of the following terms:

prime₁(0)
prime₁(s₁(0))
prime₁(s₁(s₁(x0)))
prime1₂(x0, 0)
prime1₂(x0, s₁(0))
prime1₂(x0, s₁(s₁(x1)))
divp₂(x0, x1)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

PRIME₁(s₁(s₁(x))) -> PRIME1₂(s₁(s₁(x)), s₁(x))
PRIME1₂(x, s₁(s₁(y))) -> DIVP₂(s₁(s₁(y)), x)
PRIME1₂(x, s₁(s₁(y))) -> PRIME1₂(x, s₁(y))

The TRS R consists of the following rules:

prime₁(0) -> false
prime₁(s₁(0)) -> false
prime₁(s₁(s₁(x))) -> prime1₂(s₁(s₁(x)), s₁(x))
prime1₂(x, 0) -> false
prime1₂(x, s₁(0)) -> true
prime1₂(x, s₁(s₁(y))) -> and₂(not₁(divp₂(s₁(s₁(y)), x)), prime1₂(x, s₁(y)))
divp₂(x, y) -> =₂(rem₂(x, y), 0)

The set Q consists of the following terms:

prime₁(0)
prime₁(s₁(0))
prime₁(s₁(s₁(x0)))
prime1₂(x0, 0)
prime1₂(x0, s₁(0))
prime1₂(x0, s₁(s₁(x1)))
divp₂(x0, x1)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

PRIME₁(s₁(s₁(x))) -> PRIME1₂(s₁(s₁(x)), s₁(x))
PRIME1₂(x, s₁(s₁(y))) -> DIVP₂(s₁(s₁(y)), x)
PRIME1₂(x, s₁(s₁(y))) -> PRIME1₂(x, s₁(y))

The TRS R consists of the following rules:

prime₁(0) -> false
prime₁(s₁(0)) -> false
prime₁(s₁(s₁(x))) -> prime1₂(s₁(s₁(x)), s₁(x))
prime1₂(x, 0) -> false
prime1₂(x, s₁(0)) -> true
prime1₂(x, s₁(s₁(y))) -> and₂(not₁(divp₂(s₁(s₁(y)), x)), prime1₂(x, s₁(y)))
divp₂(x, y) -> =₂(rem₂(x, y), 0)

The set Q consists of the following terms:

prime₁(0)
prime₁(s₁(0))
prime₁(s₁(s₁(x0)))
prime1₂(x0, 0)
prime1₂(x0, s₁(0))
prime1₂(x0, s₁(s₁(x1)))
divp₂(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

PRIME1₂(x, s₁(s₁(y))) -> PRIME1₂(x, s₁(y))

The TRS R consists of the following rules:

prime₁(0) -> false
prime₁(s₁(0)) -> false
prime₁(s₁(s₁(x))) -> prime1₂(s₁(s₁(x)), s₁(x))
prime1₂(x, 0) -> false
prime1₂(x, s₁(0)) -> true
prime1₂(x, s₁(s₁(y))) -> and₂(not₁(divp₂(s₁(s₁(y)), x)), prime1₂(x, s₁(y)))
divp₂(x, y) -> =₂(rem₂(x, y), 0)

The set Q consists of the following terms:

prime₁(0)
prime₁(s₁(0))
prime₁(s₁(s₁(x0)))
prime1₂(x0, 0)
prime1₂(x0, s₁(0))
prime1₂(x0, s₁(s₁(x1)))
divp₂(x0, x1)

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.

PRIME1₂(x, s₁(s₁(y))) -> PRIME1₂(x, s₁(y))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PRIME1₂(x1, x2) = PRIME1₁(x2)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > PRIME1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

prime₁(0) -> false
prime₁(s₁(0)) -> false
prime₁(s₁(s₁(x))) -> prime1₂(s₁(s₁(x)), s₁(x))
prime1₂(x, 0) -> false
prime1₂(x, s₁(0)) -> true
prime1₂(x, s₁(s₁(y))) -> and₂(not₁(divp₂(s₁(s₁(y)), x)), prime1₂(x, s₁(y)))
divp₂(x, y) -> =₂(rem₂(x, y), 0)

The set Q consists of the following terms:

prime₁(0)
prime₁(s₁(0))
prime₁(s₁(s₁(x0)))
prime1₂(x0, 0)
prime1₂(x0, s₁(0))
prime1₂(x0, s₁(s₁(x1)))
divp₂(x0, x1)

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