proof of Generalized De Morgan's Laws by mathematical induction Mathematical induction: Prove the following Generalized De Morgan's Laws.
$\sim({p_1\land p_2 \land \cdots \land p_n}) \iff \sim{p_1}\lor\sim{p_2}\lor\cdots\lor\sim{p_n}$ 
My attempt:
I'll use mathematical induction for the proof: 
 If p(n) is a statement involving the natural number n such that
p(1) is true, and p(k)⇒p(k+1) for any arbitrary natural number k, then p(n) is true for any natural number n.
(1) Let n=1, then the statement is obviously true because it becomes $\sim{p_1}\iff\sim{p_1}$. The first conditioin is satisfied
(2) Let n=k $\sim({p_1 \land p_2 \land \cdots \land p_k})$
I don't know how to further develop this proof from the second condition for the mathematical induction. I know how to prove 1+2+3+...+n= n(n+1)/2 is true for every natural numbers, but the connective $↔$ makes the problem more difficult.
FYI
Theorem De Morgan's law of $\sim(p \land q) \equiv ~\sim p ~\lor \sim q$ can be proved, using truth table.
$$\begin{array}{c|c|c|c}
\lnot({ p ∧ q }) &\iff&\ \lnot{p} ∨ \lnot{q} \\\hline
FTTT& T& FFF\\
TTFF& T& FTT\\
TFFT& T& TTF\\
TFFF& T& TTT 
\end{array}$$
EDIT: Now I know how to complete the proof
back to (2) Suppose when n=k, p(k) is true. That is, $\sim({p_1\land p_2 \land \cdots \land p_k}) \iff \sim{p_1}\lor\sim{p_2}\lor\cdots\lor\sim{p_k}$ 
Let p1 ∧p2 ∧⋯∧pk=P, ∼p1 ∨∼p2 ∨⋯∨∼pk=~Q
The truth table for $\sim({P\land p_k+1}) \iff\ \sim Q\lor \sim {p_k+1}$ 
$$\begin{array}{c|c|c|c}
\lnot({ P ∧ p_k+1 }) &\iff&\ \lnot{~Q} ∨ \lnot{pk+1} \\\hline
FTTT& T& FFF\\
TTFF& T& FTT\\
TFFT& T& TTF\\
TFFF& T& TTT 
\end{array}$$
Therefore, $\sim({p_1\land p_2 \land \cdots \land p_n}) \iff \sim{p_1}\lor\sim{p_2}\lor\cdots\lor\sim{p_n}$ for any natural number n.

First $n = 2$
$\lnot (p_1 \lor p_2)  \Leftrightarrow (\lnot p_1 \land \lnot p_2)$
Can be easily proven with a truth table.
Assume
$ \lnot (p_1 \lor p_2 \lor \cdots \lor p_n)  \Leftrightarrow (\lnot p_1 \land \lnot p_2 \land \cdots \land \lnot p_n) $   , $\forall n \in \mathbb{N}$
Now $n \rightarrow n+1$
$ \lnot (p_1 \lor p_2 \lor \cdots \lor p_n \lor p_{n+1}) $
$ \Leftrightarrow \lnot ((p_1 \lor p_2 \lor \cdots \lor p_n) \lor p_{n+1}) $
Use $n = 2$:
$ \Leftrightarrow \lnot(p_1 \lor p_2 \lor \cdots \lor p_n) \land \lnot p_{n+1}$
Use assumption now:
$ \Leftrightarrow (\lnot p_1 \land \lnot p_2 \land \cdots \land \lnot p_n) \land \lnot p_{n+1}$
$ \Leftrightarrow \lnot p_1 \land \lnot p_2 \land \cdots \land \lnot p_n \land \lnot p_{n+1}$
