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Prove the following Generalized Distributive Laws.

p∧(${q_1}$ $\lor$ ${q_2}$ $\lor$...$\lor$ ${q_n}$) ⇔ (p$\land$ ${q_1}$)∨(p$\land$ ${q_2}$)∨....∨(p$\land$ ${q_n}$)

My attempt

When n=1, the statement is obviously true since p $\land$ ${q_1}$ ⇔ p $\land$ ${q_1}$

Suppose n=k is true, that is p∧(${q_1}$ $\lor$ ${q_2}$ $\lor$...$\lor$ ${q_k}$) ⇔ (p$\land$ ${q_1}$)∨(p$\land$ ${q_2}$)∨....∨(p$\land$ ${q_k}$)

The next step is proving the statement holds true when n=k+1, but I don't know how to develop this proof further. Unlike the proof of Generalized De Morgan's Laws by mathematical induction, substituting terms doesn't seem possible.

FYI! The following truth table shows that p∧(q∨r) and (p∧q)∨(p∧r) have the same truth values in each of all logical possibilities.

$$\begin{array}{c|c|c|c} \space p\space\space q\space\space r\space \space q \space \lor \space r \space\space\space\space p \land q \space\space\space p \space \land \space r & p\land(q\space \lor \space r) \space (p\land q)\lor(p\land r) \\\hline T\space T\space T\space\space\space T\space\space\space\space\space\space\space T\space\space\space\space\space\space\space\space T & T \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space T \\ T\space T\space F\space\space \space T\space\space\space\space\space T\space\space\space\space\space\space\space\space F& T \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space T\\ T\space F\space T\space\space\space T\space \space\space\space \space F\space\space\space\space\space\space\space\space T& T \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space T\\ T\space F\space F \space\space \space F\space \space\space\space \space F \space\space\space\space\space\space\space\space F & F \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space F\\ F\space T\space T\space\space \space T\space \space\space\space \space F\space\space\space\space\space\space\space\space F & F \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space F\\ F\space T\space F\space\space\space T\space\space\space\space\space F \space\space\space\space\space\space\space\space F & F \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space F\\ F\space F\space T\space\space\space T\space\space\space\space F \space\space\space\space\space\space\space\space F & F \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space F\\ F\space F\space F\space\space\space F\space\space\space\space F\space\space\space\space\space\space\space\space F& F\space\space\space\space \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space F\\ \end{array}$$

Community
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buzzee
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1 Answers1

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What you use is associativity and break it up in two steps using the distributive law for less terms twice:

$$p\land(q_1\lor q_2 \lor \cdots \lor q_n \lor q_{n+1}) \\ \Leftrightarrow p\land((q_1\lor q_2 \lor \cdots \lor q_n) \lor q_{n+1}) \\ \Leftrightarrow (p\land(q_1\lor q_2 \lor \cdots \lor q_n)) \lor (p \land q_{n+1}) \\ \Leftrightarrow ((p\land q_1)\lor (p\land q_2) \lor \cdots \lor (p \land q_n)) \lor (p \land q_{n+1}) \\ \Leftrightarrow (p\land q_1)\lor (p\land q_2) \lor \cdots \lor (p \land q_n) \lor (p \land q_{n+1}) $$

skyking
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