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 \mathrm{q} \wedge \sim \mathrm{p}) \\ $

[By associative and commutative laws]

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

[By idempotent law]

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

$ \begin{aligned} &\equiv(p \wedge(\sim p \vee q)) \rightarrow q \\\\ &\equiv \sim(p \wedge(\sim p \vee q)) \vee q \\\\ &\equiv \sim p \vee(\sim(\sim p \vee q)) \vee q \\\\ &\equiv \sim p \vee(p \wedge \sim q) \vee q \quad d n f \\ \end{aligned} $