1. **Problem statement:** Given the functional dependencies $F=\{AB\to E; AG\to I; BE\to I; E\to G; GI\to H\}$, prove that $AB \to GH$. 2. **Recall the closure definition:** To prove $AB \to GH$, we find the closure of $AB$, denoted $(AB)^+$, and check if it contains $G$ and $H$. 3. **Calculate $(AB)^+$ step-by-step:** - Start with $(AB)^+ = \{A, B\}$. - From $AB \to E$, add $E$: $(AB)^+ = \{A, B, E\}$. - From $E \to G$, add $G$: $(AB)^+ = \{A, B, E, G\}$. - From $AG \to I$, since $A$ and $G$ are in the closure, add $I$: $(AB)^+ = \{A, B, E, G, I\}$. - From $GI \to H$, since $G$ and $I$ are in the closure, add $H$: $(AB)^+ = \{A, B, E, G, I, H\}$. 4. **Conclusion:** Since $G$ and $H$ are in $(AB)^+$, we have $AB \to GH$. Final answer: **$AB \to GH$ is proven.**