Show that each conditional statement is a tautology without using truth tables.

1. [(p-->q) ^ (q-->r)] --> (p-->r)

2. [p ^ (p-->q)]-->q

My book isn't exactly making it clear on how to do problems like these. Very frustrating. Can someone lead me in the right direction? All I know is that I'm supposed to use Logical Equivalences.

Re: Show that each conditional statement is a tautology without using truth tables.

Quote:

Originally Posted by

**lamentofking** 2. [p ^ (p-->q)]-->q.

$\displaystyle \begin{array}{*{20}{c}}{\left[ {p \wedge \left( {p \to q} \right)} \right]}\\{\left[ {p \wedge \left( {\neg p \vee q} \right)} \right]}\\{\left( {p \wedge \neg p} \right) \vee \left( {p \wedge q} \right)}\\{\left( {p \wedge q} \right)}\\q\end{array}$

Re: Show that each conditional statement is a tautology without using truth tables.

Quote:

Originally Posted by

**Plato** $\displaystyle \begin{array}{*{20}{c}}{\left[ {p \wedge \left( {p \to q} \right)} \right]}\\{\left[ {p \wedge \left( {\neg p \vee q} \right)} \right]}\\{\left( {p \wedge \neg p} \right) \vee \left( {p \wedge q} \right)}\\{\left( {p \wedge q} \right)}\\q\end{array}$

Thanks. So would the first one be:

[(p-->q) ^ (q-->r)] --> (p-->r)

[(~p v q) ^ (~p v r)]

(~p v q ^ ~p) ^(~p v q v r)

.....

Okay I'm stuck. lol. I guess I missed the first step huh. I don't know what I'm doing wrong.

Re: Show that each conditional statement is a tautology without using truth tables.

For the hypothetical syllogism (as opposed to categorical syllogism), it may be easier to prove that the negation is false. Note that ~(p -> q) = p /\ ~q. Take the negation of the whole formula and apply this rule to the outer implication (third from the left) and to p -> r. Replace the first two implications with disjunctions. Then use distributivity.