Difference between revisions of "De Morgan's Laws"
(Proof of both of De Morgan's laws using truth tables.) |
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In fact, '''all''' dual operators will interchange upon negation. So we can also say that for any [[proposition]] P, <math>\neg\forall x \: P(x) \equiv \exists x \:\neg P(x)</math>, because <math>\forall</math> is dual with <math>\exists</math>. Also, <math>\neg\exists x \:P(x) \equiv \forall x \:\neg P(x)</math>. | In fact, '''all''' dual operators will interchange upon negation. So we can also say that for any [[proposition]] P, <math>\neg\forall x \: P(x) \equiv \exists x \:\neg P(x)</math>, because <math>\forall</math> is dual with <math>\exists</math>. Also, <math>\neg\exists x \:P(x) \equiv \forall x \:\neg P(x)</math>. | ||
+ | |||
+ | ==Proof== | ||
+ | Any two propositions <math>p</math> and <math>q</math> have four possible combinations of truth values. We can therefore prove that two propositions stated in terms of <math>p</math> and <math>q</math> are equivalent by proving that they hold the same value in each of the four cases. | ||
+ | |||
+ | In the following truth table, <math>\top</math> indicates "true" and <math>\bot</math> indicates "false". | ||
+ | |||
+ | {| class="wikitable" | ||
+ | |- style="text-align: center;" | ||
+ | | <math>p</math> || <math>q</math> || <math>\neg p</math> || <math>\neg q</math> || <math>p \vee q</math> || <math>p \wedge q</math> || <math>\neg(p\vee q)</math> || <math>\neg p \wedge \neg q</math> || <math>\neg(p\wedge q)</math> || <math>\neg p \vee \neg q</math> | ||
+ | |- style="text-align: center;" | ||
+ | | <math>\top</math> || <math>\top</math> || <math>\bot</math> || <math>\bot</math> || <math>\top</math> || <math>\top</math> || <math>\bot</math> || <math>\bot</math> || <math>\bot</math> || <math>\bot</math> | ||
+ | |- style="text-align: center;" | ||
+ | | <math>\top</math> || <math>\bot</math> || <math>\bot</math> || <math>\top</math> || <math>\top</math> || <math>\bot</math> || <math>\bot</math> || <math>\bot</math> || <math>\top</math> || <math>\top</math> | ||
+ | |- style="text-align: center;" | ||
+ | | <math>\bot</math> || <math>\top</math> || <math>\top</math> || <math>\bot</math> || <math>\top</math> || <math>\bot</math> || <math>\bot</math> || <math>\bot</math> || <math>\top</math> || <math>\top</math> | ||
+ | |- style="text-align: center;" | ||
+ | | <math>\bot</math> || <math>\bot</math> || <math>\top</math> || <math>\top</math> || <math>\bot</math> || <math>\bot</math> || <math>\top</math> || <math>\top</math> || <math>\top</math> || <math>\top</math> | ||
+ | |} | ||
+ | |||
+ | Hence <math>\neg(p\vee q) \Longleftrightarrow \neg p \wedge \neg q</math> and <math>\neg(p\wedge q) \Longleftrightarrow \neg p \vee \neg q</math>. | ||
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Latest revision as of 18:53, 19 February 2022
De Morgan's Laws are two very important laws in the fields of set theory and boolean algebra.
Statement
For any two mathematical statements and , . The dual of this statement is also true, that is, . Also, for any two sets and , . Again, the dual is true, for .
In fact, all dual operators will interchange upon negation. So we can also say that for any proposition P, , because is dual with . Also, .
Proof
Any two propositions and have four possible combinations of truth values. We can therefore prove that two propositions stated in terms of and are equivalent by proving that they hold the same value in each of the four cases.
In the following truth table, indicates "true" and indicates "false".
Hence and .
This article is a stub. Help us out by expanding it. ._.