Show that the following argument form is valid. p --> q q --> r ∴ p --> r

We have to prove (p-->q) and (q-->r) implies (p-->r)

$((p\to q)\land(q\to r))\to(p\to r)\equiv$

$using(a\to b)\equiv(\neg a\lor b)$

$\equiv\neg((p\to q)\land(q\to r))\lor(p\to r)\equiv$

$using \ \neg(a\land b)\equiv(\neg a\lor \neg b)$

$(\neg(p\to q)\lor\neg(q\to r))\lor(p\to r)\equiv$

$using(a\to b)\equiv(\neg a\lor b)$

$\neg(\neg p\lor q)\lor\neg(\neg q\lor r)\lor(\neg p\lor r)\equiv$

$using \neg(a\lor b)\equiv(\neg a\land \neg b)$

$(\neg\neg p\land \neg q)\lor(\neg\neg q\land\neg r)\lor(\neg p\lor r)\equiv$

$using \neg\neg a\equiv a$

$( p\land \neg q)\lor( q\land\neg r)\lor\neg p\lor r\equiv$

$using \ commutative \ law$

$( p\land \neg q)\lor\neg p\lor( q\land\neg r)\lor r\equiv$

using\ associative\ law\

$(( p\land \neg q)\lor\neg p)\lor(( q\land\neg r)\lor r)\equiv$

$using\ distributive\ law$

$(( p\lor\neg p)\land (\neg q\lor\neg p))\lor(( q\lor r)\land(\neg r\lor r))\equiv$

$using\ a\lor \neg a \equiv T$

$(T\land (\neg q\lor\neg p))\lor(( q\lor r)\land T)\equiv$

$using \ T\land a\equiv a$

$(\neg q\lor \neg p)\lor(q\lor r)\equiv$

$using\ associative\ law$

$(\neg q\lor q)\lor(\neg p\lor r)\equiv$

$using\ a\lor \neg a \equiv T$

$T\lor(\neg p\lor r)\equiv$

$using\ T\lor a\equiv T$

$T$

Q.E.D.