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:

eq(0, 0) → true
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(s(x), s(y)) → eq(x, y)
or(true, y) → true
or(false, y) → y
union(empty, h) → h
union(edge(x, y, i), h) → edge(x, y, union(i, h))
reach(x, y, empty, h) → false
reach(x, y, edge(u, v, i), h) → if_reach_1(eq(x, u), x, y, edge(u, v, i), h)
if_reach_1(true, x, y, edge(u, v, i), h) → if_reach_2(eq(y, v), x, y, edge(u, v, i), h)
if_reach_2(true, x, y, edge(u, v, i), h) → true
if_reach_2(false, x, y, edge(u, v, i), h) → or(reach(x, y, i, h), reach(v, y, union(i, h), empty))
if_reach_1(false, x, y, edge(u, v, i), h) → reach(x, y, i, edge(u, v, h))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
eq(s(x), s(y)) →⁺ eq(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 [ ].

(2) BOUNDS(n^1, INF)