Propositional Logic Tautology Proof

Question

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.

Answer 1

    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)

Answer 2

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.

Answer 3

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.