Kenneth Rosen Edition 7 Exercise 1.3 Question 25 (Page No. 35)

Question

Show that $(p \rightarrow r) \vee (q \rightarrow r)$ and $(p \wedge q) \rightarrow r$ are logically equivalent.

Answer 1

$(p\rightarrow r)\vee(q\rightarrow r)\equiv(\sim p\vee r)\vee(\sim q\vee r)$

Now we can change propositional operator into boolean operator for easy calculation

$\wedge\equiv\cdot,\vee\equiv +$

$(p\rightarrow r)\vee(q\rightarrow r)\equiv(\bar p+ r)+(\bar q+ r)$

$(p\rightarrow r)\vee(q\rightarrow r)\equiv(\bar p+\bar q)+ r$

we can write like this

$(p\rightarrow r)\vee(q\rightarrow r)\equiv(\sim p\vee \sim q)+ r$

$(p\rightarrow r)\vee(q\rightarrow r)\equiv\sim(p\wedge q)\vee r$

Hence

${\color{Magenta}{(p\rightarrow r)\vee(q\rightarrow r)\equiv(p\wedge q)\rightarrow r} }$

Answer 2

..........

Answer 3