A specific example that with a negative answer comes from Karatsuba multiplication in which one multiplies two polynomials $$(a_0 + a_1x)(b_0 + b_1x) =: c_0 + c_1x + c_2x^2$$ $$= (a_0b_0) + (a_0b_1 + a_1b_0)x + (a_1b_1)x^2$$ $$= (a_0b_0) + (a_0b_0 + (a_0+a_1)(b_0 + b_1) + a_1b_1)x + (a_1b_1)x^2$$ If we take $\mathbb{F}_2$-coefficients (or $2$-coefficients, in your notation), this can be viewed as three Boolean functions $2^4 \to 2^1$, or their concatenation $2^4 \to 2^3$.

Karatsuba's trick allows one to turn the four multiplications and one addition into three multiplications and four additions. If you're counting all operations, and not just multiplications, it looks like you have increased from five to seven operations. But if you use this trick recursively, or take the $a_i,b_j$ to be polynomials themselves, then the reduction in multiplication wins out, and you can save overall operations.

I don't have a proof that two of the coefficients cannot be optimized without the third, but the reason I believe it is the case is because the Karatsuba trick exploits the fact that our coefficients of interest live in a rank-3 subspace of a 4-dimensional space.

A specific example that has a negative answer comes from Karatsuba multiplication in which one multiplies two polynomials $$(a_0 + a_1x)(b_0 + b_1x) =: c_0 + c_1x + c_2x^2$$ $$= (a_0b_0) + (a_0b_1 + a_1b_0)x + (a_1b_1)x^2$$ $$= (a_0b_0) + (a_0b_0 + (a_0+a_1)(b_0 + b_1) + a_1b_1)x + (a_1b_1)x^2$$ If we take $\mathbb{F}_2$-coefficients (or $2$-coefficients, in your notation), this can be viewed as three Boolean functions $2^4 \to 2^1$, or their concatenation a single function $2^4 \to 2^3$.

Karatsuba's trick allows one to turn the four multiplications and one addition into three multiplications and four additions. If you're counting all operations, and not just multiplications, it looks like you have increased from five to seven operations. But if you use this trick recursively, or take the $a_i,b_j$ to be polynomials themselves, then the reduction in multiplication wins out, and you can save overall operations.

I don't have a proof that two of the coefficients cannot be optimized without the third, but the reason I believe it is the case is because the Karatsuba trick exploits the fact that our coefficients of interest live in a rank-3 subspace of a 4-dimensional space.