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:

table → gen(s(0))
gen(x) → if1(le(x, 10), x)
if1(false, x) → nil
if1(true, x) → if2(x, x)
if2(x, y) → if3(le(y, 10), x, y)
if3(true, x, y) → cons(entry(x, y, times(x, y)), if2(x, s(y)))
if3(false, x, y) → gen(s(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
10 → s(s(s(s(s(s(s(s(s(s(0))))))))))

Rewrite Strategy: INNERMOST

(1) DecreasingLoopProof (EQUIVALENT transformation)

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