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) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:

[prime₁, prime1₂] > false > rem₂
[prime₁, prime1₂] > [s₁, and₂] > rem₂
[prime₁, prime1₂] > true > rem₂
[prime₁, prime1₂] > not₁ > rem₂
[prime₁, prime1₂] > divp₂ > 0 > rem₂
[prime₁, prime1₂] > divp₂ > =₂ > rem₂

Status:

=₂: multiset
not₁: [1]
true: multiset
divp₂: multiset
false: multiset
s₁: multiset
prime1₂: [2,1]
and₂: [2,1]
prime₁: [1]
0: multiset
rem₂: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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)

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE