Discrete math - conjunction or disjunction in this case? 
Teacher asks students if they did the homework on their own (everyone either did it on their own or copied it). He gets the following answers:
  
  
*
  
*Andy: Everyone didn't do their homework on their own.
  
*Barry: If Andy and Cindy did their homework then David copied it
  
*Cindy: Andy did it on her own or Barry did it on his own
  
*David: Barry did it on his own or Cindy did it on her own.
  
  
*
  
*Q1: Can everyone tell the truth?
  
*Q2: Can they all lie?
  
*Q3: The ones that did their homework on their own are talking the truth, the ones that copied it are lying. Who copied their homework?
  
  

The biggest question I have here is: What do we say for Andy?  Do we use the disjunction or a conjunction?
I did it this way $ \lnot A \lor \lnot B \lor \lnot C \lor \lnot D $  and I'm wondering if it's the right way or is the right way to use a conjunction?
 A: The phrase "Everyone didn't do their homework on their own" is most likely intended to mean "All students didn't do their homework on their own." That is, the conjunction option: $$\neg A \land \neg B \land \neg C \land \neg D$$
For, if the disjunction was meant, we would have far more logical means to express this, e.g.:


*

*Someone didn't do their homework on their own

*Not everyone did their homework on their own 

A: Ffor Q3 you need to solve:
$ A \leftrightarrow ( \lnot A \lor  \lnot b \lor  \lnot C \lor  \lnot D ) $
$ B \leftrightarrow ( ( A \land C) \to  \lnot D ) $
$ C \leftrightarrow (  A \lor B ) $
$ D \leftrightarrow (  B \lor C ) $
Good luck
A: Some background:
Andy's statement (Everyone didn't do their homework on their own) is better expressed using universal quantification, hence First Order Logic. Now, we know that the universal quantifier is equivalent to a conjunction of all elements of the universe:
$$ \forall x (\varphi) \equiv \bigwedge_{ a \in U} [\overline{a}/x]\varphi $$
Your Answer:
Since your universe is composed of 4 entities alone, namely, Andy, Barry, Cindy and David, it means that
$$ \forall x (\varphi) \equiv [\overline{a}/x]\varphi \wedge [\overline{b}/x]\varphi \wedge [\overline{c}/x]\varphi \wedge [\overline{c}/x]\varphi \wedge [\overline{d}/x]\varphi $$
Now, translating to your propositional glossary, this is the same as saying:

$$¬A\wedge¬B\wedge¬C\wedge¬D$$

