The statements you call A) and B) are unfortunately not true in general.
Below are two explicit counter-examples that violate them, followed by an explanation of why no “time–length versus profit” rule of the type you proposed can hold.

Notation that will be used:

Lk = log(xk) (the samples that are actually averaged)
Mk = (1/n) Σi=0…n-1 Lk-i (the n-point moving average of the logs)
t* = first instant that fulfils L t* ≤ M t* − d
t** = first instant after t* with L t** ≥ M t** (crossing from below)
Δ = t** − t*
winnum = L t** − L t*

1. A counter-example with Δ < n but winnum ≤ 0

Choose

n = 4 d = 5
t : 1 2 3 4 5 6 7
L t : 10 10 10 0 −1 −1 −0.5

• At t = 4 we have
M4 = (10+10+10+0)/4 = 7.5, so L4 = 0 ≤ 7.5 − 5 ⇒ t* = 4.

• For t = 5 and 6 we stay below the average:
M5 = 4.75, L5 = -1 < 4.75 M6 = 2, L6 = -1 < 2.

• At t = 7 we cross from below:
M7 = (0 −1 −1 −0.5)/4 = -0.625, L7 = -0.5 ≥ -0.625 ⇒ t** = 7.

Hence Δ = 7 − 4 = 3 < n = 4 but winnum = -0.5 − 0 = -0.5 ≤ 0.

This contradicts both directions of your claimed equivalence.

2. A counter-example with Δ ≥ n but winnum > 0

A very small example already works with n = 2.

Take n = 2, d = 0.5 and the log sequence

t : 1 2 3 4
L t : 2 0 −2 2

• t = 2: M2 = (2+0)/2 = 1, 0 ≤ 1 − 0.5 ⇒ t* = 2.

• t = 3: M3 = (0 −2)/2 = -1, -2 < -1 (still below).

• t = 4: M4 = (−2 +2)/2 = 0, 2 ≥ 0 ⇒ t** = 4.

Now
Δ = 4 − 2 = 2 = n and winnum = 2 − 0 = +2 > 0,

again violating your rule (you predicted a non-positive winnum whenever Δ ≥ n).

Why the “rule” cannot be correct

Write the moving-average sum as
S k = Σi=0…n-1 L k-i so that Mk = S k / n.

The “cross from below” condition at t** is equivalent to

 (n−1) L t**  ≥  S t** − L t**                 (1)

While L t* must satisfy

 L t*  ≤  S t* / n  − d                        (2)

Between t* and t** the n-sample window keeps sliding, continuously discarding the very large pre-t* values and adding new (possibly very low) values. By choosing those earlier values large enough and the in-between values low enough you can make S t** drop to practically any number you like—independently of the size of L t* itself. Consequently L t** can be forced to cross the new smaller average while still being smaller, equal to, or larger than L t*. The sign of winnum therefore cannot be determined solely from whether t** − t* is smaller than, equal to, or larger than n.

Indeed, as the two explicit sequences above show, every combination

Δ < n   with   winnum > 0 or ≤ 0,  
Δ ≥ n   with   winnum > 0 or ≤ 0

can be realised just by a suitable choice of the samples.

Literature

Because the asserted relation is false, it does not appear as a known result in probability theory, time–series analysis, DSP or any other field I am aware of. What is well known is that an un-weighted moving average does not impose monotone behaviour on the individual samples inside the window; only the average itself is smoothed.

So the short answers are:

A) Yes – the sequences above are counter-examples.
B) No – the statement is not a known theorem; in fact it is incorrect.