Are there some techniques for checking whether a statement implies another without truth tables? Are there some techniques for checking whether a statement implies another without truth tables?
For example, I was asked whether $P\Longrightarrow P_{1}$ given the following statements:
$$P: [p \land (q \land r)]\lor \neg[p \lor (q \land r)],$$
$$P_{1}: [p \land (q \lor r)]\lor \neg[p\lor(q\lor r)].$$
What I did was to find the logical equivalences of the negations of $\neg[p \lor (q \land r)]$ and $\neg[p\lor(q\lor r)]$, then tested some values, with the negations done, it was slightly easier to see what was happening as I drew 0's and 1's under $p,q,r$. I find that for $p:0,q:1,r:0$, $P$ and $P_{1}$ does not have the same truth statement, so $P\nRightarrow P_{1}$, yet it was still a tedious process.

You could write a proof instead. Suppose $P$. 
If $p\land q\land r$, then $p\land(q\lor r)$. Therefore $P_1$ tautologically.
Otherwise $\neg[p \lor (q \land r)]\Leftrightarrow \neg p \land \neg(q\land r)$. Then $p$ being true contradicts $P$, so $P\Rightarrow P_1$ in that case. In the case that $p$ is false, $P$ reduces to $\neg q \lor \neg r$ and $P_1$ reduces to $\neg q \land \neg r$. If $\neg q\not= \neg r$, then $P$ and not $P_1$, so $P\not\Rightarrow P_1$; otherwise $P$ and $P_1$ are both true or both false, so $P\Rightarrow P_1$.
Therefore $P\Rightarrow P_1$ exactly when $p$ is true or $q=r$, and $P\not\Rightarrow P_1$ exactly when $p$ is false and $q\not=r$.
