Let me state the reasoning that I used to prove that the following compound statement, composed of two primitive propositions p and q, is a tautology:
(p $\land$ (p $\Rightarrow$ q)) $\Rightarrow$ q
Now, the above statement is going to be false iff q is false and (p $\land$ (p $\Rightarrow$ q)) is true. The latter statement is true iff p is true and p $\Rightarrow$q is true. Since q is false, p must be false as well (ex falso quodlibet). Thus, we have shown that p is both true and false. That is an absurdity as p is a proposition and propositions cannot be both true and false. Hence, the compound statement cannot be false. It will always be true and so, we have proven that it is a tautology.
So, the above paragraph constitutes the reasoning that I have used. The book that I'm using, Fundamentals of Mathematics by Bernd Schroder, proves the tautology by looking at the individual cases. Of course, that's rather understandable but I'm wondering if my proof above counts as a valid one. The thing that I'm afraid of is that it looks like a Proof by Contradiction, which is proof method that is shown to be valid through the use of another tautology. I will require assistance in verifying the validity of the reasoning that I have used above.
MASSIVE EDIT BELOW:
I've been looking at the problem and trying to find other ways to prove the given compound statement. I was thinking of writing it out as disjunction of conjunctions. Let me write out the proof that I was considering as follows:
p $\Rightarrow$ q = ( p $\land$ q ) $\lor$ ( $\lnot$p $\land$ q ) $\lor$ ( $\lnot$p $\land$ $\lnot$q )
Now, we can see that the first two conjunctions can be simplified using the distributive law of Boolean Algebra. Of course, I'm also going to be using the commutative law for the conjunction operator. Hence:
( q $\land$ p ) $\lor$ ( q $\land$ $\lnot$p ) = q $\land$ ( q $\lor$ $\lnot$q).
Now, the disjunction of a proposition and its negation is going to be trivially true. Hence, the original implication can be reduced to the following statement:
p $\Rightarrow$ q = q $\lor$ ( $\lnot$p $\land$ $\lnot$q )
By the distributive law, we have:
p $\Rightarrow$ q = ( q $\lor$ $\lnot$p ) $\land$ ( q $\lor$ $\lnot$q )
Once again, we can see that the disjunction of a proposition and its negation is trivially true. Hence, we conclude that:
p $\Rightarrow$ q = ( q $\lor$ $\lnot$p )
Now, we consider the full antecedent of the compound statement from before, which is:
p $\land$ ( q $\lor$ $\lnot$p )
We can simplify it using the distributive law:
p $\land$ ( q $\lor$ $\lnot$p ) = ( p $\land$ q ) $\lor$ ( p $\land$ $\lnot$p )
The latter is an absurdity so it can be disregarded as well. Hence, we have:
( p $\land$ q ) $\Rightarrow$ q
Now, we can simplify this once more so that it is a series of conjunctions and disjunctions. After doing that, we should get:
( ( p $\land$ q ) $\Rightarrow$ q ) = ( q $\lor$ $\lnot$ ( p $\land$ q ) )
Using De Morgan's Law, we conclude that:
( ( p $\land$ q ) $\Rightarrow$ q ) = ( q $\lor$ $\lnot$p $\lor$ $\lnot$q ) = ( q $\lor$ $\lnot$q ) $\lor$ ( $\lnot$p )
We can see that the disjunction is ALWAYS true because there is a tautology embedded in it as a statement. The truth values of p and q are entirely irrelevant. Since the statement above is equivalent to the compound statement we wanted to study, we have shown that that statement is a tautology.
Once again, the above is a separate proof that I have come up with and I'd greatly appreciate any form of criticism to the proof above (assuming it is a proof).