If a negation is outside the parenthesis like this, $\sim (p\wedge \sim q)$, does it negate the whole thing. I mean like how if you have like $3(5+3)$ do you like you distributive property or am I over thinking this.
Asked
Active
Viewed 2.0k times
1
-
Yes it does. The negation is a unary operator and a parenthesized expression is equivalent to one statement. If you want to take it inside (distribute it) you have to use De Morgan's Law. A simple example can be seen here. – typesanitizer Sep 09 '14 at 01:54
2 Answers
2
You have $\sim(P \wedge \sim Q) \iff \sim P \vee Q $ by DeMorgan's law. The parens dictate the order of ops.
ncmathsadist
- 49,383
-
1I'd have written $(\sim P)\vee Q$, just to make it clear that the $\sim$ applies only to the $P$. Someone who asks this kind of question might not be clear about that, I would think. – Michael Hardy Sep 09 '14 at 01:12
0
$$ \sim(A\wedge B\wedge C\wedge D\wedge\cdots) \text{ is equivalent to }(\sim A)\vee(\sim B)\vee(\sim C)\vee (\sim D)\vee\cdots $$ $$ \sim(A\vee B\vee C\vee D\vee\cdots) \text{ is equivalent to }(\sim A)\wedge(\sim B)\wedge(\sim C)\wedge (\sim D)\wedge\cdots $$ These are de Morgan's laws, named after Augustus de Morgan, a 19th-century Englishman.
You can see that these are true by using truth tables.
Sometimes one sees them applied to union and intersection of sets where instead of $\sim$ meaning logical negation, it means complementation, and they're still called de Morgan's laws then.
A similar thing applies to existential and universal quantifiers: $$ \sim\exists x\ P\text{ is equivalent to } \forall x\ (\sim P) $$ $$ \sim\forall x\ P\text{ is equivalent to }\exists x\ (\sim P) $$