Solution to Show that the following argument form is valid. p --> q q --> r ∴ … - Sikademy
Author Image

Archangel Macsika

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

The Answer to the Question
is below this banner.

Can't find a solution anywhere?


Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

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



Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-5-stid-8-sqid-4104-qpid-2803