Divisibility labeling on a boolean lattice and positive Euler totient Let $B_n$ be the rank $n$ boolean lattice (i.e. the subset lattice of $\{1,2, \dots , n  \}$). Let $\hat{0}$ and $\hat{1}$ be the minimum and the maximum of $B_n$. Let $f: B_n \to \mathbb{N}$ be a labeling satisfying:    


*

*$f(\hat{0}) = 1$

*$a < b \Rightarrow f(a) < f(b)$ and $f(a) \mid f(b)$       

*$  f(a)f(b) \le f(a \vee b)f(a \wedge b)$, $\forall a,b \in B_n$ 


Let $\varphi(f) := (-1)^n\sum_{a \in B_n} (-1)^{|a|}f(a)$, with $|a|$ the cardinal of $a$ as subset of  $\{1,2, \dots , n  \}$.  
Example: For $B_3$ labeled as below we have $\varphi(f) = 30-6-10-15+2+3+5-1 = 8$.

Question: Is it true in general that $\varphi(f) > 0$ ?    
Examples: the divisor lattice of a square-free integer is a labeled boolean lattice satisfying the properties above, and $\varphi(f)$ is exactly the Euler totient of this integer, so that $\varphi(f) > 0$.
If we label a boolean interval of finite groups $[H,G]$ with $f(K) = |K:H|$, the labeling satisfies the properties above using the product formula, and $\varphi(f)$ is exactly the number of cosets $Hg$ such that $\langle Hg \rangle = G$, so that $\varphi(f)>0$ by Ore's theorem.  
Checking:  We have checked $B_3$ by SAGE for $f(\hat{1})<10^6$.   
Remark: it is immediate for $B_1$, and for $B_2$, take $x_1 = f(\{1,2\})$,  $x_2 = f(\{1\})$,  $x_3 = f(\{2\})$ and  $x_4 = f(\emptyset)$,  then $x_2 + x_3 \le \frac{x_1}{2} +  \frac{x_1}{2} < x_1+x_4$.
 A: It's false in general, there is a counter-example of rank $6$. 
Consider the labeling $f$ on a boolean lattice of rank $n$ such that for every maximal chain $$\hat{0} = a_0 < a_1< \cdots <a_n = \hat{1}$$ we have $$(f(a_0), f(a_1), \dots , f(a_n)) =(1,k,k^2, \dots, k^{n-2}, k^{n-2}x, k^{n-2}x^2)$$
with $k,x \in \mathbb{N}_{\ge 2}$ and $k \le x $. Then, all the properties are satisfied. But $$ \varphi(f)=  k^{n-2}x^2 - n k^{n-2}x + \sum_{r=0}^{n-2} {n \choose r} (-1)^{n-r} k^r $$ $$ = k^{n-2}x^2 - n k^{n-2}x + (k-1)^n  - k^n + nk^{n-1}$$  $$ = (k-1)^n + k^{n-2}(x-k)(x+k-n)$$
For $x = k+1$ and $n = x+k+1$, we have  $\varphi(f) = (k-1)^{2k+2} - k^{2k}<0$ for $k \le 4.$   
So with $k=2$, we have a counter-example of rank $6$, whose labeling on every maximal chain is $$(1,2,4,8,16,48,144)$$ and $\varphi(f) = -15$.   
Now, the natural questions are the following:


*

*Is $6$ the smallest rank for the existence of a labeling $f$ (as in the question) with $\varphi(f) \le 0$?  

*Is $\varphi(f)$ always nonzero for such $f$? (see this post)

