The Runtime Complexity (innermost) 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
and(true, y) → y
and(false, y) → false
size(empty) → 0
size(edge(x, y, i)) → s(size(i))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
reachable(x, y, i) → reach(x, y, 0, i, i)
reach(x, y, c, i, j) → if1(eq(x, y), x, y, c, i, j)
if1(true, x, y, c, i, j) → true
if1(false, x, y, c, i, j) → if2(le(c, size(j)), x, y, c, i, j)
if2(false, x, y, c, i, j) → false
if2(true, x, y, c, empty, j) → false
if2(true, x, y, c, edge(u, v, i), j) → or(if2(true, x, y, c, i, j), and(eq(x, u), reach(v, y, s(c), j, j)))

Rewrite Strategy: INNERMOST

(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)