Conjunctive/disjunctive normal form

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SUMMARY

The discussion focuses on converting the logical expression p <-> q into its disjunctive normal form (DNF) and conjunctive normal form (CNF). The DNF is represented as (p ∧ q) ∨ (~p ∧ ~q), while the CNF is expressed as (~p ∨ q) ∧ (~q ∨ p). DeMorgan's laws are applied to achieve these transformations, confirming the validity of the derived forms.

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  • Understanding of propositional logic
  • Familiarity with logical operators (AND, OR, NOT)
  • Knowledge of DeMorgan's laws
  • Ability to manipulate logical expressions
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  • Learn about truth tables for validating DNF and CNF
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Let's say I have p<-> q.. how do I get the disjunctive and conjunctive normal form ? I think I'd have to apply DeMorgan to get one of them. I don't know. Can anybody please explain this to me?
 
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p<->q = (p [tex]\wedge[/tex] q) [tex]\vee[/tex] (~p [tex]\wedge[/tex] ~q)
=(p [tex]\wedge[/tex] q) [tex]\vee[/tex] (~p [tex]\wedge[/tex] ~q) [tex]\vee[/tex] (p [tex]\wedge[/tex] ~p) [tex]\vee[/tex] (q [tex]\wedge[/tex] ~q)
=(~p [tex]\vee[/tex] q) [tex]\wedge[/tex] (~q [tex]\vee[/tex] p)

So, the form in red is DNF, while the form in blue is CNF.
 

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