Conditional Statements/ Implication statements within a proof specifically linear algebra I am sort of new to the mathstackexchange so excuse me for any mistakes that I make while writing this post.
I've been working on proofs and ran into a conflicted view of how to prove conditional statements. From what I know in a conditional statement proof for example, $P \implies Q$ we would suppose the antecedent i.e. P and try to prove the consequence, Q. And this is the confusing part for me as follows: let $P = J \implies K$ this means $P \implies Q$ is now equal to $(J \implies K) \implies Q$. The statement $(J \implies K) \implies Q$ in a proof would mean if we can somehow get the assertion J to be true then K would be true as well. And then try to prove for consequence Q naturally. 
The following was the question I was trying to prove and saw the solutions for.
Question: Suppose that $\{v_1, v_2\}$ is linearly independent and $w_1 = av_1+bv_2$ and $w_2 = cv_1+dv_2$ for some scalars $a,b,c,d \in \mathbb{R}.$
Claim: If $\{w_1, w_2\}$ is linearly independent then the matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $ is invertible."
Linear independence can be written as a form of conditional statement i.e. let $J : xw_1+yw_2 = 0$ and $K: x=y=0$ and $J \implies K$ would be the definition of linear independence. Let $Q :$ A is invertible. Then the claim is equivalent to $(J \implies K) \implies Q$.
This means if we can get somehow get $J$ to be true then we can conclude that $K$ is true as well. But in the solution that I looked at doesn't really obey of how I thought proofs work... I'm sure that the solution is correct but don't really understand how it's right because of the problem that I am having at the moment! Here is the solution:
SOLUTION: Suppose that $\{w_1, w_2\}$ is linearly independent. Then the equation $xw_1+yw_2=0$ has only the trivial solution and by the chain of equivalence we established... and continues...
The above is how exactly the solution was phrased and there are some missing pieces that I didn't write because the other parts irrelevant to the question I have.
So my question is when did they ever confirm $x=y=0$? Because they are simply stating the $J$ statement when that is the CONDITION...  $J$ is the assertion that they have to accomplish to say that the equation $xw_1+yw_2=0$ has the trivial solution, i.e. $x=y=0$. Proving the question itself given the above fully flows, but that one step I stated above just does not make sense to me! Any clarification or help will be very appreciated, thank you very much in advance.
 A: First of all, I am upvoting because I think your analysis and consequent question are both excellent.  Your summary that (J implies K) is asserted to imply Q is accurate.  However, you are getting bogged down by the logic.  Imagine writing all of the premises at the top of a page, writing the conclusion at the bottom of the page, then simultaneously working from the bottom upwards and from the top downwards.
From the bottom upwards, you have to prove that ad-bc is not = 0.  Now assume that $v_1 = (1,0)$ and $v_2 = (0,1)$ so that $v_1, v_2 \in \mathbb{R^2}.$
Now try to use the premises, including the fact that $w_1, w_2$ are linearly independent to establish that ad-bc is not = 0.  While doing this, use the proof given as a guide.  In effect, as a preliminary step, you are constructing your own proof in the simpler case of $v_1 = (1,0)$ and $v_2 = (0,1).$ 
Once you have your own personal proof of the simpler case (that you consider valid), extend your personal proof to cover the general case of $v_1, v_2$ being linearly independent in any vector space.  Once you have done this, compare your personal proof with the proof given in the book.  Then go back and consider how you came to be bogged down, even though your analysis was accurate. 
Per javacoder's request : explicit proof 
---------------------------------------- 
Proof by contradiction.  Assume A not invertible 
$\;\Rightarrow ad = bc.$ 
Suppose that one of $\{a,d,b,c\} = 0.\;\;$ 
WLOG $\;a=0 \;\Rightarrow\;$ one of $\{b,c\} = 0.$ 
$a=0, b=0 \;\Rightarrow\; w_1=0 \;\Rightarrow\; w_1, w_2\;$ not linearly independent. 
$a=0, c=0 \;\Rightarrow\; w_1=bv_2, w_2=dv_2 \;\Rightarrow\; w_1, w_2\;$ not linearly independent. 
Thus, each of $\{a,b,c,d\}\;$ is non-zero. 
$dw_1 - bw_2 = adv_1 + bdv_2 - bcv_1 - bdv_2 = 0 \;\Rightarrow\; w_1, w_2\;$ not linearly independent.
Per javacoder's 2nd request : answer enhanced re his followup comment 
---------------------------------------- 
First of all, I'm going to assume that you have verified that my logic is valid and that I have therefore supplied a valid proof.  If this is not true, please comment further.
I normally attack a problem like this with the following steps.


*

*Write down the premises at the top of the (first) page. 

*If necessary, use a separate page to decipher the premises, so that I'm clear on their meaning. 

*Write down the conclusion at the bottom of the (first) page, again ensuring that I am clear on its meaning. 

*As I indicated above, try to work upwards from the bottom (e.g. invertible iff ad = bc) and try to work downwards from the top (e.g. what does the linear independence of $w_1, w_2$ algebraically signify). 

*If I can connect my downwards from the top and upwards from the bottom analysis, then I'm done.

*Assuming not, what is stopping me (e.g. what if some of $a,b,c,d$ are equal to zero)?

*Further, am I having trouble because the premises are too general?  If so, I will go to a fresh piece of paper, and start over, examining a specific case [e.g. what if $v_1, v_2 \in \mathbb{R^2},\; v_1 = (1,0), v_2 = (0,1)$]. 

*If I am still stuck, with step 7, it is time to post on a math forum, with my query including a summary of all my research/analysis/questions.  If I am successful with step 7, then I will try to extrapolate to the general premise.  In other words, what is causing the assertion to be true in the specific case?  How can I establish that this cause (or a similar cause) will hold in the general case? 

*Again, If I am stuck, it is time to post on a math forum.


When I said "...getting bogged down by the logic..." I meant that (in effect) you (seemed to) successfully accomplish steps 1,2 and 3 (above), and then stopped.
