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:

prime(0) → false
prime(s(0)) → false
prime(s(s(x))) → prime1(s(s(x)), s(x))
prime1(x, 0) → false
prime1(x, s(0)) → true
prime1(x, s(s(y))) → and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) → =(rem(x, y), 0)

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

prime(0) → false
prime(s(0)) → false
prime(s(s(x))) → prime1(s(s(x)), s(x))
prime1(x, 0) → false
prime1(x, s(0)) → true
prime1(x, s(s(y))) → and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) → =(rem(x, y), 0)

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

prime(0) → false
prime(s(0)) → false
prime(s(s(x))) → prime1(s(s(x)), s(x))
prime1(x, 0) → false
prime1(x, s(0)) → true
prime1(x, s(s(y))) → and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) → =(rem(x, y), 0)

The set Q consists of the following terms:

prime(0)
prime(s(0))
prime(s(s(x0)))
prime1(x0, 0)
prime1(x0, s(0))
prime1(x0, s(s(x1)))
divp(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:

PRIME(s(s(x))) → PRIME1(s(s(x)), s(x))
PRIME1(x, s(s(y))) → DIVP(s(s(y)), x)
PRIME1(x, s(s(y))) → PRIME1(x, s(y))

The TRS R consists of the following rules:

prime(0) → false
prime(s(0)) → false
prime(s(s(x))) → prime1(s(s(x)), s(x))
prime1(x, 0) → false
prime1(x, s(0)) → true
prime1(x, s(s(y))) → and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) → =(rem(x, y), 0)

The set Q consists of the following terms:

prime(0)
prime(s(0))
prime(s(s(x0)))
prime1(x0, 0)
prime1(x0, s(0))
prime1(x0, s(s(x1)))
divp(x0, x1)

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

PRIME(s(s(x))) → PRIME1(s(s(x)), s(x))
PRIME1(x, s(s(y))) → DIVP(s(s(y)), x)
PRIME1(x, s(s(y))) → PRIME1(x, s(y))

The TRS R consists of the following rules:

prime(0) → false
prime(s(0)) → false
prime(s(s(x))) → prime1(s(s(x)), s(x))
prime1(x, 0) → false
prime1(x, s(0)) → true
prime1(x, s(s(y))) → and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) → =(rem(x, y), 0)

The set Q consists of the following terms:

prime(0)
prime(s(0))
prime(s(s(x0)))
prime1(x0, 0)
prime1(x0, s(0))
prime1(x0, s(s(x1)))
divp(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

PRIME(s(s(x))) → PRIME1(s(s(x)), s(x))
PRIME1(x, s(s(y))) → DIVP(s(s(y)), x)
PRIME1(x, s(s(y))) → PRIME1(x, s(y))

The TRS R consists of the following rules:

prime(0) → false
prime(s(0)) → false
prime(s(s(x))) → prime1(s(s(x)), s(x))
prime1(x, 0) → false
prime1(x, s(0)) → true
prime1(x, s(s(y))) → and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) → =(rem(x, y), 0)

The set Q consists of the following terms:

prime(0)
prime(s(0))
prime(s(s(x0)))
prime1(x0, 0)
prime1(x0, s(0))
prime1(x0, s(s(x1)))
divp(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof

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

PRIME1(x, s(s(y))) → PRIME1(x, s(y))

The TRS R consists of the following rules:

prime(0) → false
prime(s(0)) → false
prime(s(s(x))) → prime1(s(s(x)), s(x))
prime1(x, 0) → false
prime1(x, s(0)) → true
prime1(x, s(s(y))) → and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) → =(rem(x, y), 0)

The set Q consists of the following terms:

prime(0)
prime(s(0))
prime(s(s(x0)))
prime1(x0, 0)
prime1(x0, s(0))
prime1(x0, s(s(x1)))
divp(x0, x1)

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.


PRIME1(x, s(s(y))) → PRIME1(x, s(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PRIME1(x1, x2)  =  x2
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Precedence:
trivial

Status:
s1: [1]

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ PisEmptyProof

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

prime(0) → false
prime(s(0)) → false
prime(s(s(x))) → prime1(s(s(x)), s(x))
prime1(x, 0) → false
prime1(x, s(0)) → true
prime1(x, s(s(y))) → and(not(divp(s(s(y)), x)), prime1(x, s(y)))
divp(x, y) → =(rem(x, y), 0)

The set Q consists of the following terms:

prime(0)
prime(s(0))
prime(s(s(x0)))
prime1(x0, 0)
prime1(x0, s(0))
prime1(x0, s(s(x1)))
divp(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.