Is this proof correct? I'm trying to prove a tautology without truth tables. Conclude that $(p\land q)\rightarrow p$ is a tautology without using truth tables. Here's what I did: $$(p\land q)\rightarrow p
\equiv \lnot(p\land q)\lor p$$ By De Morgan's Law: $$\lnot(p\land q)\lor p\equiv \lnot p\lor \lnot q\lor p$$ Then By the Associative Property: $$\lnot p\lor \lnot q\lor p\equiv \lnot q \lor(\lnot p\lor p)$$
After this, I specified that $\lnot p\lor p$ is always true, so by the definition of a disjunction $\lnot q \lor(\lnot p\lor p)$ must also be true. Therefore, since $(p\land q)\rightarrow p\equiv \lnot q \lor(\lnot p\lor p)$ and $\lnot q \lor(\lnot p\lor p)$ is always true, it stands to reason that $(p\land q)\rightarrow p$ is also always true.
However, my textbook simply says "If the hypothesis $p\land q$ is true,
then by the definition of conjunction, the conclusion $p$ must
also be true."
I don't understand how this concludes the statement is a tautology. Is my proof not correct? Did I go overboard?
 A: Your proof is just fine. in fact, you might be happy to know that there are equivalence principles to do those last steps purely algebraically as well:
$$(p \land q) \to p \Leftrightarrow \text{ Implication}$$
$$\neg ( p \land q) \lor p \Leftrightarrow \text{ DeMorgan}$$
$$(\neg p \lor \neg q) \lor p \Leftrightarrow \text{ Commutation}$$
$$(\neg q \lor \neg p) \lor p \Leftrightarrow \text{ Association}$$
$$\neg q \lor (\neg p \lor p) \Leftrightarrow \text{ Complement}$$
$$\neg q \lor \top \Leftrightarrow \text{ Annihilation}$$
$$\top$$
A: $(p\land q)\rightarrow p$
OP: Did I go overboard? The book simply states
"If the hypothesis $p\land q$ true, then by the definition of conjunction, the conclusion $p$ must also be true."
Answer: You showed in detail that it is a tautology, but it can get tedious being so exact. In fact, the reason for defining the implication symbol is that it makes life simpler. You could make logical arguments using only negation, disjunction and conjunction, but it is cumbersome - you aren't a computer. Also, disjunction is preferred to using the 'exclusive OR' - it is a more natural way of expressing statements.
Using logical implications allows you to see a 'flow'. You can usually just jump on the premise and assume it is true, and of course you can also use the contraposition method. This "implication flow" is a comfortable way of handling tautologies and proving theorems.
Book Technique: $(p\land q)\rightarrow p\;\;\;\;\;\;$ (1)
Assume the hypothesis $p\land q\;\;\;\;\;\;\;\;$  (2) is TRUE.
We must show that $p$ is also TRUE.
But for the conjunction (2) to be TRUE, $p$ must be TRUE. 
So (1) is a tautology.
Note that you also have these tautologies,
$(p\land q\land r)\rightarrow p$
$(p\land q\land r)\rightarrow q$
$(p\land q\land r)\rightarrow r$
Eventually this 'class of tautologies' can all be taken for granted.
A: The textbook is correct as the truth value of the implication will always be true, by the way implication is defined, and hence the statement is a tautology.
The way you went about showing the statement is a tautology is also correct. 
Another way of proving that it is a tautology without resorting to truth tables:
Assume that there are truth value assignments to $p$ and $q$ which make our implication false (hence assuming it is not a tautology). An implication $A\to B$ is false iff $A$ is true and $B$ is false. 
Therefore we would need to have $p\land q$ true yet $p$ is false. This is a contradiction and so no such truth value assignments can be made to $p$ and $q$ hence $(p\land q)\to p$ must be a tautology.
A: In fact, your proof is correct. You can use a truth table to ascertain that you are right:
$p$ | $q$ | $(p∧q)$ | $(p∧q)→p$
0 | 1 | _ 0  _ |___ 1
1 | 0 | _ 0  _ |___ 1
0 | 0 | _ 0  _ |___ 1
1 | 1 | _ 1  _ |___ 1
You can see that any combination of $p$ and $q$ gives TRUTH.
A: Theorem
$(p \wedge q) \to p$.
Proof
Assume $p \wedge q.$ Then $p.$ Thus $(p \wedge q) \to p.$
This is how the formula is proven using natural deduction. I've just written the proof using text instead of inference rules.
