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    Discrete Math Help needed

    here's the theorem that i'm tryin to prove...it's a basic one, i just wanna be sure that i am doin this right...symbols that i'm usin here, "-" represents "NOT" and "==" represents "IS EQUIVALENT TO"...and of course, "=" is equals, and ":=" represents "is substitued by"...so here it is

    --------------------------------------------
    Theorem: -- p == q == p == --q

    Using Distribution of - over ==: - (p == q) == -p == q, for "--p == q" from the theorem; and using Leibniz for the rest of the theorem to carry it down as it is:

    = - (- (p == q)) == p == --q

    Using Symmetry of == on "p == --q" from the last line: p == q == q == p, where q:= --q; and leibneiz on the rest:

    = - (- (p == q)) == --q == p

    for "--q == p" from the last line, using Distribution of - over ==: - (p == q) == -p == q, where p:= q, q:= p; and using Leibniz for the rest to carry it down as it is:

    = - (- (p == q)) == - (- (p == q))

    Reflexivity of == states: p == p; here p:= -( - (p ==q)), therefore intitial theorm is true.

    --------------------------------------------

    **phew**

    can someone look over it, c if it's correct? and can someone think of any other ways to prove the theorem??

    thanx in advance

    ------------------
    If I'm dreaming, never let me wake. If I'm awake, never let me sleep.

    #2
    http://www3.pak.org/gupshup/smilies/confused.gif
    Yaad aaiy jo dard ki raat may, jis ki baat ho apni har baat may.

    Comment


      #3
      no one??

      ------------------
      If I'm dreaming, never let me wake. If I'm awake, never let me sleep.

      Comment


        #4
        ok here's another one...a lil more complex, but still not 2 bad...just not sure bout the end though...it's a long one...if anyone has another way of proving this, plz let me know...

        symbols that i'm usin here, "-" represents "Not/Negation" ; "--" represents "Double Negation"; "==" represents "Is Equivalent to" ; "=" is "equals" ; ":=" represents "is substitued by" ; "!==" represents "is not equivalent to"...so here it is:

        --------------------------------------------
        Theorem: ( ( p !== q ) !== r ) == ( p !== ( q !== r ) )

        on ( ( p !== q ) !== r ) from the theorem, applying Definition of !==: ( p !== q ) == - ( p == q ) where p:= ( p !== q ) and q:= r ;and Leibniz on the rest:

        = - ( ( p !== q ) == r ) == ( p !== ( q !== r ) )

        on ( p !== ( q !== r ) ) from the above statement, applying Definition of !==: ( p !== q ) == - ( p == q ) where q:= ( q !== r ) ;and Leibniz on the rest:

        = - ( ( p !== q ) == r ) == - ( p == ( q !== r ) )

        on ( p !== q ) from the above statement, applying Definition of !==: ( p !== q ) == - ( p == q ) ; and Leibniz on the rest:

        = - ( - ( p == q ) == r ) == - ( p == ( q !== r ) )

        on ( q !== r ) from the above statement, applying Definition of !==: ( p !== q ) == - ( p == q ) where p:= q, q:= r ; and Leibniz on the rest:

        = - ( - ( p == q ) == r ) == - ( p == - ( q == r ) )

        on - ( p == q ) from the above statement, applying Distributivity of - over ==: - ( p == q ) == - p == q ; and Leibniz on rest:

        = - ( - p == q == r ) == - ( p == - ( q == r ) )

        on - ( q == r ) from the above statement, applying Distributivity of - over ==: - ( p == q ) == - p == q where p:= q, q:= r; and Leibniz on rest:

        = - ( - p == q == r ) == - ( p == - q == r )

        on ( p == - q ) from the above statement, applying Symmetry of ==: p == q == q == p where q:= - q; and Leibniz on rest:

        = - ( - p == q == r ) == - ( - q == p == r )

        on - ( - p == q == r ), applying Distributivity of - over ==: - ( p == q ) == - p == q where p:= - p, q:= q == r; and Leibniz on rest:

        = -- p == q == r == - ( - q == p == r )

        on - ( - q == p == r ), applying Distributivity of - over ==: - ( p == q ) == - p == q where p:= - q, q:= p == r; and Leibniz on rest:

        = -- p == q == r == -- q == p == r


        From Previous Theorem, -- p == p, therefore:

        = p == q == r == -- q == p == r

        on -- q, applying Theorem: -- p == p where p == q; and Leibniz on rest:

        = p == q == r == q == p == r

        on ( q == p ) from the above statement, applying Symmetry of ==: p == q == q == p; and Leibniz on rest:

        = p == q == r == p == q == r

        Now here is what i don't know...

        Using the Axiom of Associativity of ==: ( p == ( q == r ) ) == ( ( p == q ) == r ) -----> Is it a valid step??

        or


        Using Theorem of Relexivity of ==: p == p where p:= p == q == r -----> this should be valid for sure
        --------------------------------------------

        *phew*

        comeon...where are all the math experts here?? not that an expert is needed for this proof http://www3.pak.org/gupshup/smilies/biggrin.gif


        ------------------
        If I'm dreaming, never let me wake. If I'm awake, never let me sleep.

        Comment


          #5
          sorry dm bhai


          all i see that you are messing with the "p" key "r" key and the "equal" key on your keyboard http://www3.pak.org/gupshup/smilies/biggrin.gif http://www3.pak.org/gupshup/smilies/hehe.gif

          Life became all Gray! But NOW i have decided to paint it all over again.

          I Tawt I Taw A Puddy Tat

          Comment


            #6
            unfortunately i dont remember Discrete Maths much http://www3.pak.org/gupshup/smilies/frown.gif 'cuz its really been a long time (around 5+ yrs)

            but for your second proof, wouldnt it be easy if you had used DeMorgan's law instead...
            which i believe is:
            -(p AND/OR q) == -p AND/OR -q
            and then making use of that Contrapositive logical equivalency....i.e:
            p AND/OR q == -q AND/OR -p

            ah! cant remember it exactly http://www3.pak.org/gupshup/smilies/frown.gif

            Comment


              #7
              Hang on there brother
              i'll help you in a couple of years http://www3.pak.org/gupshup/smilies/tongue.gif

              Comment


                #8
                I did all this crap and I thank God I forgot it all.

                Sorry DM, couldn't help you.

                [This message has been edited by sambrialian (edited May 12, 2002).]

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