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Math Help - Logic

  1. #1
    Junior Member
    Joined
    Sep 2006
    Posts
    27

    Logic

    How do I completely solve this problem? I have created truth tables for a↓b and a|b but I don't understand the rest of the question.


    Here it is:
    Connectives ↓ (Pierces's arrow, nor) and | (Scheffer's stroke, nand) are defined by a↓b := 7(a v b), and a|b := 7(a ^b). Express each of these connectives in terms of 0 and →. Construct truth tables for ↓ and |. Show that each of the connectives 7, ^ and v can be expressed in terms of the connective ↓, and in terms of the connective |.

    *the 7 is to negate....
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  2. #2
    marcmtlca
    Guest

    answer?

    7a=7(ava)=a↓a

    a^b=7(7(a^b))=7(7av7b)=7a↓7b=(a↓a)↓(b↓b)

    avb=7(7(avb))=7(7a^7b)=7a|7b=(a↓a)|(b↓b)


    I used DeMorgan's laws up there btw. and the fact that double negatives cancel.

    Marc
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