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If I want to proof the following by contradiction: $$P\implies Q.$$ Say $R$ is a contradiction$(R\equiv\bot)$. By constructing $$\begin{array}{l} (P\land\lnot Q)\implies R.\\ \lnot R.\\ \hline\therefore\lnot(P\land\lnot Q).\end{array}$$ Since $$\lnot(P\land\lnot Q)\iff\lnot P\lor Q,$$ which implies $$P\implies Q.$$

Is the above correct? Is the above relate to the Law of Excluded Middle?

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    Just to clarify your question, $P,Q$ are some propositions that you want to prove the implication for, for some other math problem. You use a somewhat complicated logic, and you want to verify that the logic is sound? – 2017-02-16
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    @ZacharySelk Yes, I means the logical structure itself and I want to know whether these steps are correct. – 2017-02-16
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    Sure, that works! – 2017-02-16

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Yes, your argument looks correct.