The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)
divides(x, y) → div(x, y, y)
div(0, 0, z) → true
div(0, s(x), z) → false
div(s(x), 0, s(z)) → div(s(x), s(z), s(z))
div(s(x), s(y), z) → div(x, y, z)
prime(x) → test(x, s(s(0)))
test(x, y) → if1(gt(x, y), x, y)
if1(true, x, y) → if2(divides(x, y), x, y)
if1(false, x, y) → true
if2(true, x, y) → false
if2(false, x, y) → test(x, s(y))

Rewrite Strategy: FULL

(1) CpxTrsToCpxRelTrsProof (BOTH BOUNDS(ID, ID) transformation)

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

gt(s(x), 0) → true
gt(0, y) → false
gt(s(x), s(y)) → gt(x, y)
divides(x, y) → div(x, y, y)
div(0, 0, z) → true
div(0, s(x), z) → false
div(s(x), 0, s(z)) → div(s(x), s(z), s(z))
div(s(x), s(y), z) → div(x, y, z)
prime(x) → test(x, s(s(0)))
test(x, y) → if1(gt(x, y), x, y)
if1(true, x, y) → if2(divides(x, y), x, y)
if1(false, x, y) → true
if2(true, x, y) → false
if2(false, x, y) → test(x, s(y))

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
gt(s(x), s(y)) →⁺ gt(x, y)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [x / s(x), y / s(y)].
The result substitution is [ ].

(4) BOUNDS(n^1, INF)