Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(6) Obligation:

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.

(7) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(8) Obligation:

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

PRIME1(x, s(s(y))) → PRIME1(x, s(y))

R is empty.
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.

(9) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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

(10) Obligation:

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

PRIME1(x, s(s(y))) → PRIME1(x, s(y))

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

(11) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• PRIME1(x, s(s(y))) → PRIME1(x, s(y))
  The graph contains the following edges 1 >= 1, 2 > 2

(12) TRUE