1. The problem involves understanding the composition of operators $r_l$ and $r_m$ acting on elements $F$, $G$, and $B$ with given relations: $$r_l \circ r_m (F) = G$$ $$r_l \circ r_l (F) = B$$ and the statement: $$r_A \circ r_m (G) = F$$ 2. We want to find the answer or relationship implied by these equations. 3. From the first equation, applying $r_l$ after $r_m$ on $F$ gives $G$. 4. The third equation states that applying $r_A$ after $r_m$ on $G$ returns $F$. 5. This suggests that $r_A \circ r_m$ is the inverse operation of $r_l \circ r_m$ on $F$, meaning: $$r_A \circ r_m (G) = F = (r_l \circ r_m)^{-1} (G)$$ 6. From the second equation, applying $r_l$ twice on $F$ gives $B$: $$r_l \circ r_l (F) = B$$ 7. Without additional information about the operators $r_l$, $r_m$, and $r_A$, or the nature of $F$, $G$, and $B$, the best we can conclude is that $r_A \circ r_m$ acts as an inverse to $r_l \circ r_m$ on $F$, and $r_l$ squared on $F$ yields $B$. Final answer: The operator $r_A \circ r_m$ is the inverse of $r_l \circ r_m$ on $F$, and $r_l$ composed with itself on $F$ equals $B$.