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

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

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

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

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

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

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

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

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

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

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

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

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

by Veteran

+1 vote