Kenneth Rosen Edition 7 Exercise 1.3 Question 23 (Page No. 35)

Question

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

Answer 1

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Answer 2

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

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

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

$(p\rightarrow r)\wedge(q\rightarrow r)\equiv(\bar {p}+r)\wedge(\bar{q}+r)$

$(p\rightarrow r)\wedge(q\rightarrow r)\equiv\bar {p}.\bar{q}+\bar{p}.r+r.\bar{q}+r.r$

$(p\rightarrow r)\wedge(q\rightarrow r)\equiv\bar {p}.\bar{q}+\bar{p}.r+r.\bar{q}+r$

$(p\rightarrow r)\wedge(q\rightarrow r)\equiv\bar {p}.\bar{q}+\{\bar{p}+\bar{q}+1\}r$

$(p\rightarrow r)\wedge(q\rightarrow r)\equiv\bar {p}.\bar{q}+1.r$

$(p\rightarrow r)\wedge(q\rightarrow r)\equiv\bar {p}.\bar{q}+r$

we can write like this 

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

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

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

Hence

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

Answer 3