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:

cond1(true, x, y, z) → cond2(gr(y, z), x, y, z)
cond2(true, x, y, z) → cond2(gr(y, z), p(x), p(y), z)
cond2(false, x, y, z) → cond1(and(eq(x, y), gr(x, z)), x, y, z)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x
eq(0, 0) → true
eq(s(x), 0) → false
eq(0, s(x)) → false
eq(s(x), s(y)) → eq(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
cond2(true, x, s(x20191_3), 0) →⁺ cond2(true, p(x), x20191_3, 0)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [x20191_3 / s(x20191_3)].
The result substitution is [x / p(x)].

(2) BOUNDS(n^1, INF)