A **"general"** Boolean (combinatoiral) circuit is a labeled (with the labels: AND, OR, NOT, IN, OUT), directed, acyclic graph, that satisfies:

1. fan-in=2 for the AND and OR nodes  
2. fan-n=1 for the NOT nodes  
3. fan-in=0 for the IN nodes  
4. fan-out=0 to exactly one node (the OUT node)  
5. Unbounded fan-out to the rest of the nodes (but the OUT node)

A **monotone** circuit is a Boolean circuit with 0 vertices labeled as "NOT".

The size of a circuit is the number of "gates" (vertices with labels "AND", "OR" or "NOT") it contains.

In Yuval's answer [here](https://cs.stackexchange.com/questions/124040/is-it-assumed-that-lower-bounds-on-the-size-of-monotone-circuits-apply-to-genera) I've learned of two examples (Tardos function and bipartite perfect matching) where it has been proven that monotone circuits admit greater size than general Boolean circuits, but I cannot get the intuition, as I don't have any concrete small size example in hand.

Hence, my question is: could you please supply me with an example of a **small** (say, up to 10-20 gates) non monotone circuit such that any equivalent monotone circuit has greater size?

**Edit**

I guess that the smallest-size circuit computes the 3-Clique decision problem, as this is the smallest size where we can exploit fast matrix multiplication for the k-Clique problem (where non-monotone circuits may have smaller size than their equivalent monotone circuits, as I mentioned before).

Since the key part of exploiting fast matrix multiplication is (roughly): does $X^2$ contain any non-zero?
Hence, I guess that a circuit that computes this decision problem is the minimal-size one. So, it is equivalent to $$\bigvee_{i,j,k\leq n}{(x_{ij}\wedge x_{jk})}\equiv\bigvee_{i,j\leq n}\left(x_{ij}\wedge{\left(\bigvee_{k\leq n} x_{jk}\right)} \right)$$.

If it is true, we just need to find some small $n$ (preferably, the smallest $n$), and some non-monotone function of size $\lneq n^2\cdot (n+1)=n^3+n^2$, which is the size of RHS, which is probably the minimal-size monotone circuit that computes this function.

Now, smallest size for Strassen is $n=4$, so it admits $n^2=4^2=16$ variables, but I guess that simpler manipulation than Strassen could do the desired on smaller $n$, and with simpler manner.

To sum up, I almost get it, but still do not have an explicit simple and small example in hand.