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:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
and(x, true) → x
and(x, false) → false
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(0) → 0
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)
Log(x) → -(Log'(x), I(0))
Val(L(x)) → x
Val(N(x, l, r)) → x
Min(L(x)) → x
Min(N(x, l, r)) → Min(l)
Max(L(x)) → x
Max(N(x, l, r)) → Max(r)
BS(L(x)) → true
BS(N(x, l, r)) → and(and(ge(x, Max(l)), ge(Min(r), x)), and(BS(l), BS(r)))
Size(L(x)) → I(0)
Size(N(x, l, r)) → +(+(Size(l), Size(r)), I(1))
WB(L(x)) → true
WB(N(x, l, r)) → and(if(ge(Size(l), Size(r)), ge(I(0), -(Size(l), Size(r))), ge(I(0), -(Size(r), Size(l)))), and(WB(l), WB(r)))

Rewrite Strategy: INNERMOST

(1) DecreasingLoopProof (EQUIVALENT transformation)

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

(2) BOUNDS(n^1, INF)