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Math Help - Help proving NAND gate is equaivalent to an OR gate

  1. #1
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    Help proving Combination of NAND gates is equaivalent to an OR gate

    We have to prove that a combination of 3 NAND gates is logically equivalent to an OR gate by using a truth table. I thought this one was going to be easy but maybe I'm overthinking it a bit. I feel like I may have proved it's equal to a NOR instead of an OR. Can somebody check over this for me? From reading up on NAND gates I can see that it's just the reverse order of an OR gate (i.e if OR = 0,1,1,1 then NAND = 1,1,1,0).Assume there are 2 inputs, and we run them through 2 seperate NAND gates and then run those two outputs to a final NAND gate(3 total NAND gates).

    x y x OR y ~x ~y (x AND y) ~(x AND y) (~x OR ~y)
    0 0 0 1 1 0 1 1
    0 1 1 1 0 0 1 1
    1 0 1 0 1 0 1 1
    1 1 1 0 0 1 0 0

    Is this correct? The thing that was jumbling me up is that I was unsure of things like "is NAND written as ~(A ^ B) or (~A ^ ~ B)". Did I just prove it was equal to a NOR gate instead of an OR?
    Last edited by RogerSmith; October 1st 2013 at 10:20 PM. Reason: Had to fix an important error
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  2. #2
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    Re: Help proving NAND gate is equaivalent to an OR gate

    Hey RogerSmith.

    NAND and OR gates are not equivalent. Are you thinking of a combination of NAND gates to produce an OR or a NOR gate (which is commonly done)?
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  3. #3
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    Re: Help proving NAND gate is equaivalent to an OR gate

    Yes, I meant a combination. Maybe I should add the word combination to my first post. I did mention that there were, however, three NAND gates involved
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    Re: Help proving NAND gate is equaivalent to an OR gate

    Hint: You can obtain ~x by using NAND(x,x) or ~(x and x) = ~ x.
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    Re: Help proving NAND gate is equaivalent to an OR gate

    Quote Originally Posted by chiro View Post
    Hint: You can obtain ~x by using NAND(x,x) or ~(x and x) = ~ x.
    Right, and the same thing applies for y, but when I get to the 3rd and final NAND gate and pass ~x as 1 parameter and ~y as the other parameter, the result should be x AND y right? But how will that affect the truth table? Like will that help me prove that combination of NAND gates is equal to an OR value from my truth table?

    **EDIT***

    Ohhh I think I get what you're saying now. so if i do ~(~x and ~y), which I think is the same thing as passing the two parameters over to the last NAND gate, my results should come out to be the value of (x OR y)
    Last edited by RogerSmith; October 1st 2013 at 10:45 PM.
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