1. ## Negation Logic

Write the negation of the logical expression $(p \rightarrow \neg q) \wedge (p \vee q)$. Simplify it to the point where the negation symbol appears only in front of the variables $p$ and $q$, do not leave the negation symbol in front of a parenthetical clause.

Is the answer $(\neg q \wedge \neg p) \vee (\neg p \wedge \neg q)$?

2. Alright, I'll try to walk you though this:

¬ [(p -> ¬q) ^ (p v q)]

because of ¬(p^q)=¬p v ¬q we get

¬ (p -> ¬q) v ¬(p v q)

because of ¬(p->q)=p^¬q we get

(p ^ ¬¬q) ^ ¬(p v q)

because of ¬(p v q)=¬p^¬q we get

(p ^ ¬¬q) ^ (¬p ^ ¬q)

From there we just simplify and get:

(p ^ q) ^ (¬p ^ ¬q)

Hope that helps!

3. Hello, chiph588@!

Write the negation of: . $(p \to \sim q) \wedge (p \vee q)$
$\sim\bigg[(p\to \sim q) \wedge (p \vee q)\bigg]$

. . $= \;\sim\bigg[(\sim p \:\vee \sim q) \vee (p \vee q)\bigg]$

. . $= \;\sim(\sim p\: \vee \sim q)\: \vee \sim(p \vee q)$

. . $= \;(p \wedge q) \vee (\sim p \:\wedge \sim q)$

. . $= \;\bigg[(p \wedge q) \:\vee \sim p\bigg] \wedge \bigg[(p \wedge q) \vee \sim q\bigg]$

. . $= \;\bigg[(p \:\vee \sim p) \wedge (q \:\vee \sim p)\bigg] \wedge \bigg[(p \:\vee \sim q) \wedge (q \:\vee \sim q)\bigg]$

. . $= \;\bigg[t \wedge (q \:\vee \sim p)\bigg] \wedge \bigg[(p \:\vee \sim q) \wedge t\bigg]$

. . $= \;(q \:\vee \sim p) \wedge (p\: \vee \sim q)$

. . $= \;(\sim p \vee q) \wedge (\sim q \vee p)$

. . $= \;(p \to q) \wedge (q \to p)$

. . $= \;p \Longleftrightarrow q$