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Math Help - Prepositional logic, truth tables and Venn diagrams

  1. #1
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    Prepositional logic, truth tables and Venn diagrams

    i want to solve the following questions:

    Q.No.1

    Assume that for the truth values
    P=F, q=T, r=F
    Show that the proposition
    ~(pvq)^~(~pvr) is false.

    Q.No.2

    Formulate the arguments symbolically and test its validity using truth table.
    If you invest in the auto industry, then you get rich.
    You didn't invest in the auto industry.
    Therefore, you didn't get rich.

    Where p = "You invest in the auto industry."
    q = "You get rich."


    Q.No.3

    Illustrate the distributive law A(BC)=(AB)(AC) with Venn diagrams

    pleae help me in solving these questions.

    Azhar
    Last edited by Jhevon; April 13th 2009 at 12:39 PM. Reason: removed link
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  2. #2
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    Q.No.1

    Assume that for the truth values
    P=F, q=T, r=F
    Show that the proposition
    ~(pvq)^~(~pvr) is false.



    ~(pvq)^~(~pvr)
    F FTT F F TFTF

    false
    Last edited by princess_21; March 28th 2009 at 07:33 AM.
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  3. #3
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    Have you had a look around at wikipedia? (Eg. Logical connective - Wikipedia, the free encyclopedia) But there are many other entries which deal with this topic, such as 'propositional logic', 'boolean algebra.' and even 'truth table.'

    Formulate the arguments symbolically and test its validity using truth table.
    If you invest in the auto industry, then you get rich.
    You didn't invest in the auto industry.
    Therefore, you didn't get rich.
    Where p = "You invest in the auto industry."
    q = "You get rich."
    So, the first proposition is 'if p then q' : (  p \Rightarrow q )
    Then have a look at the corresponding truth table (implication). The other two propositions are similar to the first proposition. Remember that not p \equiv \neg p (which in the truth table is the equivalent of a truth 'value' of F(alse).)

    If you look find out what a Venn diagram is, you will be able to do the third part. Venn diagrams are very good.
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  4. #4
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    Q.No.2

    Formulate the arguments symbolically and test its validity using truth table.
    If you invest in the auto industry, then you get rich.
    You didn't invest in the auto industry.
    Therefore, you didn't get rich.

    Where p = "You invest in the auto industry."
    q = "You get rich."

    p->q
    ~p
    therefore ~q

    [(p->q)->~p]->~q
    T T FT FT
    T F FT TF
    F T TF FT
    F F TF TF


    [(p->q)->~p]->~q
    T T T FT FT
    T F F FT TF
    F T T TF FT
    F T F TF TF

    [(p->q)->~p]->~q
    T T T F FT FT
    T F F T FT TF
    F T T T TF FT
    F T F T TF TF


    [(p->q)->~p]->~q
    T T T F FT T FT
    T F F T FT T TF
    F T T T TF F FT
    F T F T TF T TF


    SO the answer is False






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  5. #5
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    i cant understand in this way becos i m new in this math.
    please help me solving this in easy and basic way using truth tables.

    Thanks
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  6. #6
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    Question 2
    Definition: A proposition is a statement that is either true or false:
    There are 2 propositions:
    1. P:= you invest in the auto industry
    2. Q:= you [become] rich

    So, each proposition can be either T(rue) or F(alse). Ie. If you did invest in the auto industry the P is True, if not, then False.

    So, there are different ways to combine two propositions, for example:
    P AND Q,  P \wedge Q
    P OR Q,  P \vee Q etc.
    Each operator has its own truth table.

    You're interested in P IMPLIES Q, which is often presented as "if P then Q"
     P \Rightarrow Q
    The truth table for P \Rightarrow Q is:

    P | Q |  P \Rightarrow Q
    -------------------
    T | T | T
    T | F | F
    F | T | T
    F | F | T

    So, now you could say, NOT P  \Rightarrow Q is T(rue) etc. Which would mean 1. (did not invest in the auto industry)  \Rightarrow 2. (becomes rich)  \equiv True
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