Taking the inverse of a statement and then substituting I'm taking a junior high/high school geometry course. We were talking about how a square is a rhombus and a rectangle, and therefore a parallelogram, but a parallelogram is not necessarily a rhombus or rectangle, much less a square. I was doing some random logic in my head and I came up with this: 
If a quadrilateral has two opposite pairs of congruent sides, it is a parallelogram 
If it does not have 2 opposite pairs o/ congruent sides, it is not a parallelogram 
If it does not not have 2 opposite pairs o/ congruent sides, it is a parallelogram 
A quadrilateral that does not have two opposite pairs of congruent sides is a trapezoid.
If a quadrilateral is a trapezoid, it is a parallelogram
Here's a simplified version, but with the same idea: 
If I am wearing orange, I can go to the Giant's game 
If I am not wearing orange, I cannot go to the Giant's game 
If I am not wearing not orange, I can go to the Giant's game. 
Something not orange is blue, so
If I am not wearing blue, I can go to the Giant's game
Obviously this is not true, so where did I go wrong?
 A: I think you are unsure of, but just learning about, converse statements. This fundamental tenant of logic is essential to getting the most out of mathematics.
Let "p" be a statement "It's sunny outside," and "q" be the statement "I'm happy."
$p \implies q$    ($p$ implies $q$) means "If it's sunny outside, then I am happy." This is true.
Now take the converse: $q \implies p$, means "If I'm happy, then it's sunny outside." Not necessarily true: a Ferrari on my driveway will make me happy even if it was overcast for an entire year.
Does that mean that if $p$ implies $q$, $q$ can never imply $p$? Not at all: consider these two statements:
$p: x^2-1=0$, and $q: x=\{-1,1\}$.
$p\implies q$ means: if $x^2-1=0$, then we know that $x=\{-1,1\}$. So this statement is true.
$q\implies p$ means: if $x=\{-1,1\}$, then squaring each element and substracting one will give us zero, i.e., $x^2-1=0$ is true.
Thus, since the implication runs both way, we say $p \iff q$, or "$p$ if and only if (a.k.a., "iff") $q$". 
Try applying these rules toward your statements. Do they clarify a few things?
