Mathematical induction: $p ∧(q_1 ∨ \cdots∨q_n) = (p∧q_1)∨\cdots∨(p∧q_n)$

Question

Prove that for every positive integer $n$, $$p ∧(q_1 ∨ \cdots∨q_n) = (p∧q_1)∨\cdots∨(p∧q_n).$$

This is a mathematical induction question. What would I test for the base case and how? Any help on how to start would be appreciated.

Answer 1

I would look at two base cases, $n=1$ and $n=2$. When $n=1$, $p \land q_1 = p \land q_1$ trivially. When $n=2$, use a truth table to prove $p \land (q_1 \lor q_2) = (p \land q_1) \lor (p \land q_2)$ (i.e. check if it is true for all 8 combinations of values for $p$, $q_1$, and $q_2$). Then you can break up $q_1 \lor\cdots\lor q_n$ into $q_1 \lor q'$ with $q'=q_2 \lor\cdots\lor q_n$ for the induction step.