## Are $\mathsf{\#P}$ problems harder than $\mathsf{NP}$ problems

I have a method to solve the $$\mathsf{\#P}$$ version of 3SAT in a way that seemingly reduces it to an $$\mathsf{NP}$$ problem. – I don’t have a formal understanding of these terms so I will just show an example here.

How many solutions does $$\phi=\left(x_1 \vee x_2 \right) \wedge \left(\neg x_1 \vee x_3 \right) \wedge \left(\neg x_2 \vee \neg x_3 \right)$$ have?

First convert to a boolean polynomial: $$\phi=\left(x_1 + x_2 -x_1x_2\right) \left(1-x_1+x_1x_3\right) \left(1-x_2x_3 \right)$$

Expand and simplify: $$\phi = x_2 – x_1x_2 + x_1x_3 – x_2x_3$$

Substitute $$x_1 = x_2 = x_3 = \frac{1}{2}$$ and multiply by $$2^n$$ where $$n$$ is the number of variables: $$|\phi| = 2^n\left(\frac{1}{2} – \frac{1}{4} + \frac{1}{4} – \frac{1}{4}\right) = 2^3\frac{1}{4} = 2$$

where $$|\phi|$$ is the number of solutions that $$\phi$$ has.

And indeed, those 2 solutions are (0,1,0) and (1,0,1).

I have proven that this works for any logical expression. Does this have any use/bearing on $$\mathsf{\#P}$$ vs $$\mathsf{NP}$$