Proving negation with multiple conditions Suppose that there is some condition A that is only true iff condition 1 and condition 2 are true. Now suppose that there is some result B and I'm trying to prove A ⇔ B. I can prove A ⇒ B, but to prove B ⇒ A I'm trying to show ¬A ⇒ ¬B. Now the problem is if I assume that condition 1 is false, and can show this implies B is false, is this enough to prove that ¬A ⇒ ¬B? Or do I also need to prove that if condition 2 is false then B is false?
 A: No, you need the second as well. Consider the situation when


*

*Condition 1 is true

*Condition 2 is false

*B is false


I'm going to call condition 1 $P$ and condition 2 $Q$ for the sake of clarity.
You know the following (101) about $A$ .
$$ A \iff P \land Q \tag{101} $$
This means that (102) is true as well.
$$ (\lnot A) \iff (\lnot P) \lor (\lnot Q) \tag{102} $$ .
You are trying to prove (103) below, the contrapositive of $B \implies A$ .
$$ (\lnot A) \implies (\lnot B) \tag{103} $$
We can subsittute for $\lnot A$ using (102), yielding (104) below.
$$ ((\lnot P) \lor (\lnot Q)) \implies (\lnot B) \tag{104} $$
Let's replace $\implies$ with disjunction and a negated left argument, yielding (105).
$$ (P \land Q) \lor (\lnot B) \tag{105} $$
Based on the passage below, you are proving (106a-b) and asking if that entails (105).

if I assume that condition 1 is false, and can show this implies B is
  false, is this enough to prove that ¬A ⇒ ¬B?

$$ (\lnot P) \implies (\lnot B) \tag{106a} $$
or equivalently
$$ P \lor (\lnot B) \tag{106b} $$
(106b) does not entail (105) . For instance, if $P$ (condition 1) is true, $Q$ (condition 2) is false, and $B$ is false, then (106b) would be true, but (105) would be false.
However, if you also knew (107a-b), then you could infer (105).
$$ (\lnot Q) \implies \lnot B \tag{107a} $$
$$ Q \lor \lnot B \tag{107b} $$
A: If you want to show $\neg A \implies \neg B$ you have to show that whenever $A$ fails, $B$ fails also.  There are two ways for $A$ to fail: condition $1$ doesn't hold, or condition $2$ doesn't hold.  So, yes, you have to deal with both cases.  
To put it another way, suppose condition $1$ holds, but condition $2$ doesn't.  Your proof has to deal with this case.  
A: 
Suppose that there is some condition A that is only true iff condition 1 and condition 2 are true.
I can prove A ⇒ B,

So you have proof that $\def\getsto{\leftrightarrow}A\getsto(C\wedge D)$ and $A\to B$.

but to prove B ⇒ A I'm trying to show ¬A ⇒ ¬B. Now the problem is if I assume that condition 1 is false, and can show this implies B is false, is this enough to prove that ¬A ⇒ ¬B?

No.
If $\lnot B$ is derivable from $\lnot C$, then $C$ must be derivable under an assumption of $B$.   However, a derivation of $C$ is insufficient to derive $C\land D$, and hence $A$ cannot be derived under an assumption of $B$ just with the given premise.

Or do I also need to prove that if condition 2 is false then B is false?

Yes. You also need $\lnot B$ to be derivable under an assumption of $\lnot D$.
