Why is $\neg (p \land q)$ different than $(\neg p \land \neg q)$?

Question

Why is $\neg (p \land q)$ different than $(\neg p \land \neg q)$?

If we let:

p: Blair is a liar

q: Bush is a liar

Then:

¬(p & q) is "Neither Bush nor Blair are liars"

which seems to be the same as:

(¬p & ¬q) "Bush is not a liar and Blair is not a liar"

But the logic tables for them are different.

$\neg$(p & q)

($\neg$p & $\neg$q)

Answer 1

"Neither Bush nor Blair are liars" can be written in two equivalent ways:

$$\lnot (p\lor q) \equiv \lnot p \land \lnot q$$

"Not both Bush and Blair are liars"

$$\lnot (p\land q) = \lnot p \lor \lnot q$$

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$\lnot(p\land q)$ means "Not both (p and q)." This means that that either $\lnot p$ or $\lnot$ q.

$\lnot(p \lor q)$ means it's not the case that (either p holds or q holds), i.e. "neither p nor q," This is equivalent, as noted above, to $\lnot$ p