Kenneth Rosen Edition 7 Exercise 1.3 Question 32 (Page No. 35)

Question

Show that $(p \wedge q) \rightarrow r $ and $(p \rightarrow r ) \wedge (q \rightarrow r)$ are not logically equivalent.

Answer 1

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

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

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

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

$(p \wedge q) \rightarrow r \equiv(\bar p+ \bar q)+ r$  --------------$>(1)$

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

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

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

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

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

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

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

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

$(p \rightarrow r ) \wedge (q \rightarrow r)\equiv\bar p.\bar q+r$ -----------$>(2)$​​​​​​​

From equation $(1)$ and $(2)$ both are not equivalent.

Answer 2

The statement "p and q implies r" is logically equivalent to "if p and q are true, then r is also true". While the statement "p implies r and q implies r" is logically equivalent to "if p is true, then r is true" and "if q is true, then r is true"

The first statement says that p and q are jointly sufficient for r, while the second statement says that p and q are individually sufficient for r. These two statements are not logically equivalent because there could be cases where p and q are both true but r is not true. For example, if p represents "It is raining" , q represents "I have an umbrella" and r represents "I will go for a walk", one could say that p and q are true but I still choose not to go for a walk which would make r false.

On the other hand, if the first statement is true, then the second statement is also true, but the converse is not true, which is why they are not logically equivalent.

Answer 3

In (p$\wedge$q)$\rightarrow$r $\equiv$ (p$\rightarrow$r)$\wedge$(q$\rightarrow$r) 

If either p=T,q=F $or$ p=F,q=T and r=F  then LHS is True and RHS is False. 

Hence, they are not logically equivalent.