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)


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 contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
QDP
              ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PRIME12(x, s1(s1(y))) -> PRIME12(x, s1(y))
Used argument filtering: PRIME12(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof
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.