Obtain disjunctive normal form of,

i) $(p \rightarrow q) \wedge(\sim p \wedge q)$

ii) $(p \wedge(p \rightarrow q)) \rightarrow q$

Solution:

(i) $(p \rightarrow q) \wedge(\sim p \wedge q) \\ $

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

$\equiv(\sim \mathrm{p} \wedge(\sim \mathrm{p} \wedge \mathrm{q}) \vee(\mathrm{q} \wedge \sim \mathrm{p} \wedge \mathrm{q}) \\ $

[Using distributive law]

$\equiv((\sim \mathrm{p} \wedge \sim \mathrm{p}) \wedge \mathrm{q}) \vee(\mathrm{q} \wedge …