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Problem 1.27(b)
Prove via mathematical induction: $$(*) \hspace{35pt} \left( \displaystyle\bigvee_{i=1}^n F_i \right) \wedge \left( \displaystyle\bigvee_{j=1}^m G_j \right) \equiv \displaystyle\bigvee_{i=1}^n \left( \displaystyle\bigvee_{j=1}^m (F_i \wedge G_j) \right).$$ Solution: Recall the (simple) distributive law (Example 1.25 in the text): $$(**) \hspace{35pt} P \wedge (Q \vee R) = (P \vee Q) \wedge (P \vee R).$$ First we will show that the formula $(*)$ holds for $m=n=1$. In this case we get the equality $$\left( \displaystyle\bigvee_{i=1}^1 F_i \right) \wedge \left( \bigvee_{j=1}^1 G_j \right) \stackrel{?}{\equiv} \bigvee_{i=1}^1 \left( \bigvee_{j=1}^1 (F_i \wedge G_j) \right),$$ simplifying each side yields $$F_1 \wedge G_1 \equiv F_1 \wedge G_1,$$ which is true, so we see that $(*)$ holds for $m=n=1$. Suppose $(*)$ holds for $n=1,2,3,\ldots,N$ and $m=1,2,\ldots,M$, i.e. assume that $$(***) \hspace{35pt} \left( \bigvee_{i=1}^N F_i \right) \wedge \left( \bigvee_{j=1}^M G_j \right) = \bigvee_{i=1}^N \left( \bigvee_{j=1}^M (F_i \wedge G_j) \right).$$ Let us prove that the formula consequently holds for $n=N+1$ (later we will also have to show it holds for $m=M+1$). Plugging in $n=N+1$ yields $$\left( \bigvee_{i=1}^{N+1} F_i \right) \wedge \left( \bigvee_{j=1}^{M} G_j \right) \stackrel{?}{\equiv} \bigvee_{i=1}^{N+1} \left( \bigvee_{j=1}^{M} (F_i \wedge G_j) \right).$$ To establish this equivalence, consider the following calculation: $$\begin{array}{ll} \left( \displaystyle\bigvee_{i=1}^{N+1} F_i \right) \wedge \left( \displaystyle\bigvee_{j=1}^M G_j \right) &\equiv \left( \left( \displaystyle\bigvee_{i=1}^N F_i \right) \vee F_{N+1} \right) \wedge \left( \displaystyle\bigvee_{j=1}^M G_j \right) \\ &\stackrel{(**)}{\equiv} \underbrace{\left( \left( \displaystyle\bigvee_{i=1}^N F_i \right) \wedge \left( \displaystyle\bigvee_{j=1}^M G_j \right) \right)}_{\text{use } (***),n=N} \vee \underbrace{\left( F_{N+1} \wedge \left( \displaystyle\bigvee_{j=1}^M G_j \right) \right)}_{\text{use } (***),N=1, \text{ and be careful!}} \\ &\equiv \left( \displaystyle\bigvee_{i=1}^N \left( \bigvee_{j=1}^M (F_i \wedge G_j) \right) \right) \vee \left( \displaystyle\bigvee_{i=N+1}^{N+1} \left( \displaystyle\bigvee_{j=1}^M F_i \wedge G_j \right) \right) \\ &\stackrel{\text{definition of }\bigvee}{\equiv} \left( \displaystyle\bigvee_{i=1}^{N+1} \left( \bigvee_{j=1}^M (F_i \wedge G_j) \right) \right). \end{array}$$ Therefore we have established $(*)$ for $n=N+1$ and $m=M$. The proof for $n=N$ and $m=M+1$ is exactly the same (swap $N$ and $M$ in the previous argument). Therefore we have established $(*)$ using strong induction.