every proposition is equivalent to a proposition in DNF
Theorem.
Given any proposition, there exists a proposition in disjunctive normal form
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which is equivalent
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to that proposition.
Proof.
Any two propositions are equivalent if and only if they determine the same truth function. Therefore, if one can exhibit a mapping which assigns to a given truth function a proposition in disjunctive normal form such that the truth function of this proposition is , the theorem follows immediately.
Then, we claim that
On the one hand, suppose that for a certain . By definition of , we have . By definition of , we have
In either case, . Since a conjunction![]()
equals if and only if each term of the conjunction equals , it follows that . Finally, since a disjunction
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equals if and only if there exists a term which equals , it follows the right hand side equals equals when the left-hand side equals .
On the one hand, suppose that for a certain . Let be any element of . Since , there must exist an index such that . For this choice of , Then we have
In either case, . Since a conjunction equals if and only if there exists a term which evaluates to , it follows that for all . Since a disjunction equals if and only if each term of the conjunction equals , it follows that the right hand side equals equals when the left-hand side equals .
∎
| Title | every proposition is equivalent to a proposition in DNF |
|---|---|
| Canonical name | EveryPropositionIsEquivalentToAPropositionInDNF |
| Date of creation | 2013-03-22 15:06:58 |
| Last modified on | 2013-03-22 15:06:58 |
| Owner | rspuzio (6075) |
| Last modified by | rspuzio (6075) |
| Numerical id | 10 |
| Author | rspuzio (6075) |
| Entry type | Theorem |
| Classification | msc 03B05 |