The Generalized Associativity Equation is given by 
$$ F(G(x,y),z)=K(x,H(y,z)),$$
where the functions $F,G,H$ and $K$ are all from $\mathbb{R}^2$ to $\mathbb{R}$. In his book "Lectures on Functional Equations and their applications" (page 329), Aczel proves that, under some differentiability assumptions, the terms in the equation can also be written as
$$f(a(x)+b(y)+c(z)).$$


**Question:** *Is there a counterexample for this statement if we do not assume any form of differentiability? To be precise, are there functions $F,G,H,K$, satisfying the generalized associativity equation, for which there are no $f,a,b,c$ such that 
$f(a(x)+b(y)+c(z))$ equals the terms in the equation?*



Counterexample for the Generalized Associativity Equation
With no continuity assumptions, you can always find such $f,a,b,c$.  

Namely, let $A, B, C$ be subsets of $\mathbb R$ with cardinality of the continuum
such that the function $(x,y,z) \to x+y+z$ is one-to-one on $A \times B \times C$.
Let $a$, $b$, $c$ be one-to-one functions from $\mathbb R$ onto $A, B, C$ respectively. Then for $s \in A + B + C$ define $f(s) = F(G(x,y),z)$ where
$(x,y,z)$ is the unique member of $\mathbb R^3$ with $a(x) + b(y) + c(z) = s$.


For example, you could take $$A = \left\{ \sum\_{j=1}^\infty 2^{-4j} x\_j \; : x \in \{0,1\}^\mathbb N\right\}$$ 
$$B = A/2 = \left\{ \sum\_{j=1}^\infty 2^{-4j-1} x\_j \; : x \in \{0,1\}^\mathbb N\right\}$$ 
$$C = A/4 = \left\{ \sum\_{j=1}^\infty 2^{-4j-2} x\_j \; : x \in \{0,1\}^\mathbb N\right\}$$ 


