Please forgive me if this question seems trivial, maybe it's because I'm teaching myself. Anyways here's the question and how I attempted it.
Q. Prove the following tautological conditional.
$$[p\implies q]\implies[(p\land r)\implies(q\land r)]$$
A. My answer
Assuming $p\implies q$ to be true, we assume $p$ to be true and therefore $q$ is true. Regardless of the truth-value of $r$, $(p\land r)\implies(q\land r)$ is true. Therefore the original statement is a tautology.
Is this the right approach? Is this enough? Should I write about the case in which $p$ is false and $q$ is false?
Here is a formal proof of the statement (which you can play with at http://proofs.openlogicproject.org/)
To prove that a statement of the form $A\to B$ is true, the most straightforward approach is to assume $A$ as a hypothesis and derive $B$ as a consequence. This statement has implications two levels deep, so we take two of those hypotheses in line 1 and 2. Our hypothesis in line 2 is a conjunction, saying that both $P$ and $Q$ are true, so we can separate them into seperate statements in lines 3 and 4. Now, since we know that $P$ and $P\to Q$, we can infer $Q$ in line 5. Since we know that $Q$ and $R$ are true (under our hypotheses, at least), we can join them together in line 6. That was the consequence we had hoped to prove, so we unwind the hypotheses back into implications in line 7 and 8 to finish the proof.