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
  ↳ 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)

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:

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)

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:

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)

Q is empty.
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
  ↳ 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)

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.


PRIME12(x, s1(s1(y))) -> PRIME12(x, 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( PRIME12(x1, x2) ) = max{0, x1 + x2 - 1}



The following usable rules [14] were oriented: none



↳ 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)

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.