+1 vote
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Show that $(p \wedge q) \rightarrow r$ and $(p \rightarrow r ) \wedge (q \rightarrow r)$ are not logically equivalent.
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$(p \wedge q) \rightarrow r \equiv\sim(p\wedge q)\vee r$

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

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.

by Boss (44.6k points)

+1 vote
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