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    We have to apply DeMorgan'd laws twice.. And I have no clue how to do it!
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    (Original post by JKITFC)
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    We have to apply DeMorgan'd laws twice.. And I have no clue how to do it!
    Why don't you consider it pairwise?

    A^c \cap B^c = (A \cup B)^c and C^c \cap D^c = (C \cup D)^c

    Now you have (A \cup B)^c \cap (C\cup D)^c = \cdots
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    (Original post by JKITFC)
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    We have to apply DeMorgan'd laws twice.. And I have no clue how to do it!
    Have a think about what de Morgan's laws say about (A \cup B)^c. Now apply the fact that  A, B are mutually exclusive, in considering  \mathbb{P}((A \cup B)^c) = 1 - \mathbb{P}(A \cup B)
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    (Original post by Zacken)
    Why don't you consider it pairwise?

    A^c \cap B^c = (A \cup B)^c and C^c \cap D^c = (C \cup D)^c

    Now you have (A \cup B)^c \cap (C\cup D)^c = \cdots
    Looks good, I got that far, didn't know the second application
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    (Original post by JKITFC)
    Looks good, I got that far, didn't know the second application
    Awesome! Couple that with Greg's hint above and all should be good.
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    (Original post by Zacken)
    Awesome! Couple that with Greg's hint above and all should be good.
    What does.. (A \cup B)^c \cap (C\cup D)^c = \cdots equal?
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    (Original post by JKITFC)
    What does.. (A \cup B)^c \cap (C\cup D)^c = \cdots equal?
    Perhaps if we write A \cup B = E and F = C \cup D that'll clarify matters: E^c \cap F^c = \cdots? Hint: De Morgan.
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    (Original post by JKITFC)
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    We have to apply DeMorgan'd laws twice.. And I have no clue how to do it!
    Why does the thread say probability? DeMorgan's laws are about formal logic.
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    (Original post by Juichiro)
    Why does the thread say probability? DeMorgan's laws are about formal logic.
    Because you can apply De Morgan laws to probabilistic events.
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    (Original post by Zacken)
    Because you can apply De Morgan laws to probabilistic events.
    Ah, that's interesting. Technically, true statements can be interpreted as statements of 100% probability so I guess it shouldn't be so surprising. Still cool though.
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    (Original post by Juichiro)
    Ah, that's interesting. Technically, true statements can be interpreted as statements of 100% probability.
    I've never really thought about it that way before! That does sound very intriguing.
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    Well you know that the LHS is saying that something other than A or B or C or D happened. and the P(something happened) = 1
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    (Original post by Juichiro)
    Ah, that's interesting. Technically, true statements can be interpreted as statements of 100% probability so I guess it shouldn't be so surprising. Still cool though.
    How do you deal with sets of measure zero?
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    (Original post by Zacken)
    I've never really thought about it that way before! That does sound very intriguing.
    Yeah, I found in an intro to logic book.

    (Original post by Gregorius)
    How do you deal with sets of measure zero?
    What does that mean in terms of propositional logic?
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    (Original post by Juichiro)

    What does that mean in terms of propositional logic?
    I am querying your assertion that true statements can be interpreted in terms of 100% probability. As probability is done in the setting of a measure defined on a sigma algebra of sets, I am wondering what you do with sets of measure zero.
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    (Original post by Gregorius)
    I am querying your assertion that true statements can be interpreted in terms of 100% probability. As probability is done in the setting of a measure defined on a sigma algebra of sets, I am wondering what you do with sets of measure zero.
    No idea. I just read it in a textbook. I know what it means in propositional logic but not in set theory.
 
 
 
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