Here is an argument from Culver’s MA thesis from 2010.

Suppose the sequence $(A_0\wedge A_1,A_2\wedge A_3,\ldots)$ is $\vee$-reducible to $A$.

Let’s look at the use of the $\vee$-truth-table reduction for computing $A_8\wedge A_9$.

We claim that after seeing only $A_0,\ldots,A_7$ we can rule out one possibility for $(A_8,A_9)$, thereby making money on $A$ and thus showing $A$ is not random.

If the use is only $A_0,\ldots,A_7$ then let’s say the reduction predicts $A_8\wedge A_9=1$. Then we can rule out the pair $(0,1)$. If the reduction predicts $0$ we can rule out $(1,1)$.

If the use is only $A_0,\ldots,A_8$ then based on us looking at $A_0,\ldots,A_7$ we can determine that the reduction predicts $1\wedge A_9=1$ or we can determine that the reduction predicts $1\wedge A_9=0$. Then we can rule out the pair $(1,0)$, or $(1,1)$, respectively.

Finally, if the reduction uses all of $A_0,\ldots,A_9$ then based on us looking at $A_0,\ldots,A_7$ we can determine that the reduction predicts $1\wedge 1=1$ and we can determine that the reduction predicts $1\wedge 0=1$ and… the reduction cannot correctly predict the whole $\wedge$ function or else $\wedge$ would be an element of the clone $\{\vee\}$.