Propositional Logic Tautology Proof I have question about a proposition that I need to prove is a tautology:
$((p \rightarrow q) \wedge (q \rightarrow r)) \rightarrow (p \rightarrow r)$
I have tried negating the first large bracket, but after a few steps I'm stuck. Should I show that the first 2 brackets are the same as $(p \rightarrow r)$ and therefore it is a tautology?
Please help me.
 A: Here is an answer through truth tables,

This is how I did it:


*

*Fill in all the variables first.

*Do the first implication from p and q

*Do the second implication from q and r

*Do the conjunction from the first and second implications

*Do the implication to the furthest right, from p and r

*Then do the remaining implication from the conjunction and the implication above.


The end result is that the proposition is true in all possible worlds, a tautology. 
A:     Suppose p->q and q->r.     Assumption for --> introduction

    p-->q                      and elimination
    q-->r                      and elim.

       Suppose p                   assumption for --> int

       q                           --> elim
       r                           --> elim
    p-->r                          --> int

(p-->q and  q-->r)-->(p-->r)           
A: I am posting my solution for others to view.
$((p→q)∧(q→r))→(p→r) \\ \equiv (p \rightarrow r) \rightarrow (p \rightarrow r) \\
\equiv T$
I think the law is called Hypothetical Syllogism.
Please correct me if its not all right.
A: The following proof uses a Fitch-style natural deduction proof checker:

To show the statement is a tautology, I will attempt to derive it without any premises. If I can do that it would be the same as putting the conditional statement into a truth table generator and showing that all valuations of the sentence letters give a true result under the top-level conditional ($\to$) connective symbol. That would make it a tautology.
Since what I want to prove is a conditional, I assume the antecedent, $(P\to Q) \land (Q\to R)$ as the start of a subproof, indented in Fitch-style natural deduction. Since this antecedent is a conjunction, I use conjunction elimination (∧E) to derive each of the conjuncts on lines 2 and 3.
Since the consequent of what I want to prove is also a conditional, I start another subproof assuming its antecedent, $P$. Given $P$, I can use modus ponens or conditional elimination (→E) on line 5 to derive $Q$ referencing lines 4 and 2. Similarly on line 6 I can derive $R$.
At this point I have a subproof assuming $P$ and deriving $R$. I can summarize that by using conditional introduction (→I) on line 7 to derive $P\to Q$.  This discharges the assumption on line 4 and closes the subproof.  Line 7 gives me the consequent of the conditional I want to derive.
I still have an undischarged subproof starting on line 1, but I note that I can discharge that assumption using conditional introduction on line 8 to derive the goal. 

The OP has the following question:

Should I show that the first 2 brackets are the same as (p→r) and therefore it is a tautology?

For the antecedent of this conditional to be the same as the consequent it would have to be a biconditional. This would show more than is required, it if were possible. (One can use a truth table generator to show that the biconditional is not a tautology.)
Rather one has to assume the antecedent and derive the consequent without using any premises. Although one is assuming the antecedent in a subproof, that assumption is later discharged closing the subproof and so one derives the goal without any premises or undischarged assumptions.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
