Is Negation of $(a \implies \neg b)$ Equivalent to $(a \implies b)$? This feels like a dumb question but I just want to make sure I'm correct. I'm reading over some homework where we need to prove some statements are equivalent and one person tried to show $a \implies b$ by assuming $a \implies \neg b$ and coming to a contradiction. This isn't actually valid, correct? 
 A: If $\lnot$(a $\rightarrow$ $\lnot$b), then a is true and $\lnot$b is false.  Thus, a is true and b is true.  So, (a $\rightarrow$ b).
However, if (a $\rightarrow$ b), it does NOT follow that $\lnot$(a $\rightarrow$ $\lnot$b), since a could be false, which renders both (a $\rightarrow$ b) and (a $\rightarrow$ $\lnot$b) true, making $\lnot$(a $\rightarrow$ $\lnot$b) false.
A: No indeed. $A \Rightarrow \neg B$ is equivalent to $\neg(A \wedge \neg\neg B) = \neg(A \wedge B)$. Hence $\neg(A \Rightarrow \neg B)$ is $A \wedge B$.
For instance: if $A$ is $x = 0$ and $B$ is $x^2 = 0$ then $A \Rightarrow \neg B$ is false and $A\wedge B$ is true.
A: Everyone's right.
But my way of thinking:
a => b is only false when a is true and b is false.  In other words it's only false 1 in 4 times.  So the negation will only be true 1 in 4 times; when a is true and b is false.  So the negation is a AND -b.
a => -b will only be false when a is true and b is true.  In other word it's only false 1 in 4 times.  So it can't be the negation of above because the negation of above can only be true 1 in 4 times.  
(a => -b will only be false when a is true and b is true so a=> -b is the same as -(a AND b).  Different thing altogether.)
A: Simple short answers. The following two are logically equivalent
$$(a \implies b) \iff  (\neg b \implies \neg a)$$ 
and also the negation of an implication is not an implication
$$\neg (a \implies b) \iff  (a \wedge \neg b)$$
If you want to prove $(a \implies b)$ is true by contradiction then you should prove that it's negation is false. Hence, you should prove that $(a \wedge \neg b)$ is false. :)
