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Prove De Morgan's LAw

  1. Jun 15, 2004 #1
    I hope someone can help me prove one of De Morgan's Law:

    (A intersection B)' = A' U B'
     
  2. jcsd
  3. Jun 15, 2004 #2

    AKG

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    I can suggest a way to prove it, although I've never formally done any of this stuff so I don't know if this approach is "acceptable." Anyways, it's a proof by contradiction. Assume the equation were false. That would mean that there exists and element, x, such that it is not in (A' U B') but is in (A intersection B)'. Now, if it is not in (A' U B'), then it is not in A' and it is not in B' (if it were in either of those, it would be in A' U B'). So, if it is not in A', it is in A, and if it is not in B', it is in B. Therefore, this element, x, is in A and it is in B, so it is in (A intersection B). Therefore, it is not in (A intersection B)'. This contradicts, the assumption, therefore the equation must be right.
     
  4. Jun 15, 2004 #3
    franz32,

    Draw truth tables for both expressions. The expressions are equivalent if and only if their truth tables are the same.
     
  5. Jun 15, 2004 #4

    Tom Mattson

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    S=A U B means, "Every element of S is either an element of A or an element of B (or of both)", and T=A ^ B means, "Every element of S is an element of both A and B".

    We can write this using set builder notation:

    A U B={x|xεA or xεB}

    Take the complement:

    (A U B)'={x|xεA or εB}'

    Taking the complement of the statement on the RHS is the same as logically negating the "or" statement inside. Thus,

    (A U B)'={x|~(xεA or xεB)}

    Now apply DeMorgan's law for logical connectives to fit this to the definition of A'^B'. You can justify the use of DeMorgan's law for logic by using truth tables.
     
  6. Jun 15, 2004 #5

    AKG

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    Proposition: [itex](A \cap B)' \subseteq (A' \cup B')[/itex]

    Proof: (by contradiction)
    Assume: [itex](A \cap B)' \not\subseteq (A' \cup B')[/itex]
    [itex]\therefore \exists x \mbox{ such that } x \in (A \cap B)',\ x \notin (A' \cup B')[/itex]
    [itex][x \notin (A' \cup B')] \Longrightarrow (x \notin A' \wedge x \notin B')[/itex]
    [itex]\Longrightarrow (x \in A \wedge x \in B)[/itex]
    [itex]\Longrightarrow [x \in (A \cap B)][/itex]
    [itex]\Longrightarrow [x \notin (A \cap B)'][/itex]
    [itex]\oplus \mbox{ (contradiction)}[/itex]​
    [itex]\therefore (A \cap B)' \subseteq (A' \cup B')\ \dots \ (1)[/itex]​
    Proposition: [itex](A' \cup B') \subseteq (A \cap B)'[/itex]

    Proof: (by contradiction)
    Assume: [itex](A' \cup B') \not\subseteq (A \cap B)'[/itex]
    [itex]\therefore \exists x \mbox{ such that } x \notin (A \cap B)',\ x \in (A' \cup B')[/itex]
    [itex][x \notin (A \cap B)'] \Longrightarrow [x \in (A \cap B)][/itex]
    [itex]\Longrightarrow (x \in A \wedge x \in B)[/itex]
    [itex]\Longrightarrow [x \notin A' \wedge x \notin B')][/itex]
    [itex]\Longrightarrow [x \notin (A' \cup B')][/itex]
    [itex]\oplus \mbox{ (contradiction)}[/itex]​
    [itex]\therefore (A' \cup B') \subseteq (A \cap B)'\ \dots \ (2)[/itex]​
    By (1) and (2), [itex](A' \cup B') = (A \cap B)'[/itex]
     
    Last edited: Jun 17, 2004
  7. Jun 15, 2004 #6
    Oh.. I see

    Well, thanks for all your helps. I got it. :biggrin:
     
  8. Jun 16, 2004 #7

    Gokul43201

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    Draw a Venn Diagram.

    It becomes painfully obvious.
     
  9. Jun 16, 2004 #8

    Tom Mattson

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    Ha! The last time I responded to a request for help in proving a set-theroetic statement, I said the same thing. But the poster said it wasn't allowed, so I did it "the hard way" this time.

    It's true though, franz32, you can show that both statements have the same Venn diagram, and are therefore equivalent.
     
  10. Jun 17, 2004 #9
    I haven't really looked at the LATEX stuff, so I'll have to do it with words.
    Let x be in (A intersection B)'.
    Then x is NOT in A intersection B
    <=> x not in A AND x not in B (use logical connectors, negations, etc)
    <=> x in A' OR x in B'
    <=> x in A' U B'

    That's one of them, hope it makes sense... you do the other one. (it's exactly the same) :smile:
     
  11. Jun 24, 2004 #10
    doing a truth table would be difficult, but a venn diagram would be the least painful way. i know you have the answer but still for further problems like this, venn is the way to go.
     
  12. Jun 24, 2004 #11

    Tom Mattson

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    It's no more difficult than the Venn diagram. Once you reduce DeMorgan's law for sets to DeMorgan's law for logic, all you have to do is show that ~(p and q) has the same truth table as (~p or ~q).

    ~(p and q)
    Since (p and q) is true only when (p,q)=(T,T), the negation is false only when (p,q)=(T,T).

    (~p or ~q)
    Since (p or q) is only false when (p,q)=(F,F), the statement (~p or ~q) is false only when (p,q)=(T,T).

    Since the two statements have the exact same truth table, they are equivalent.
     
  13. Mar 23, 2010 #12
    hello

    can some one Plz explain to me my task:

    Prove De Morgan Law with the appropriate explanation were p and q are sentence meaning (proposition)

    ~(p ^ ( it s OR i dnt knw how to type on comuter) ---(if) (~p) ^ (~q)

    and which font should i use so that i can type the symbols and if some one will prove it i will be glad
     
  14. Mar 24, 2010 #13

    Fredrik

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    The best way is to use LaTeX. Click the quote button in post #5 to see how AKG did it. (tex tags will produce slightly larger output than itex tags). You will like the LaTeX codes \lnot,\lor and \land. There are lots of lists of LaTeX symbols online that you can find using Google. I also recommend that you use the preview button to check that everything looks OK before you submit your post.
     
  15. Aug 27, 2011 #14
  16. Aug 27, 2011 #15

    phinds

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    Not to sidetrack the thread, but when did DeMorgan's Theorem become DeMorgan's Law ? I had never heard it called that but when I googled it, I got about 59,000 hits for law, 49,000 for theorem. I don't think when I learned it about 50 years ago it was every called anything but DeMorgan's Theorem.
     
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