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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Combined order from the following AFS and order.
PRIME1(x1, x2)  =  x2
s(x1)  =  s(x1)
prime(x1)  =  prime(x1)
0  =  0
false  =  false
prime1(x1, x2)  =  x2
true  =  true
and(x1, x2)  =  and
not(x1)  =  x1
divp(x1, x2)  =  x2
=(x1, x2)  =  x2
rem(x1, x2)  =  rem(x1, x2)

Recursive Path Order [RPO].
Precedence:

prime₁ > s₁ > true > [0, false]
prime₁ > s₁ > and > [0, false]
rem₂ > [0, false]

The following usable rules [FROCOS05] were oriented:

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)

(8) Obligation:

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.

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE