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Math Help - Convert Truth Table to Boolean AND-OR Expression

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    Convert Truth Table to Boolean AND-OR Expression

    Hello MHF. I am stuck on a problem that goes like this:

    Write the boolean AND-OR expression for the following:

    function x in Figure 10.17

    Convert Truth Table to Boolean AND-OR Expression-boolean19c.png

    I don't understand how to get a boolean expression from that truth table. Are true and false reversed for this problem meaning, True = 0 and False = 1? And do I just write down the expressions everywhere x = 1? So for example where a, b, and c are 0 and x=1, the expression for that part would be abc? Here's what I have so far:

    abc + abc' + ab'c + a'bc + a'bc' + (abc)'

    Is that right?
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    Re: Convert Truth Table to Boolean AND-OR Expression

    you certainly can just list the combinations that result in 1 in your truth table but I suspect they want you to simplify that.

    The first four entries can be seen to be

    $a'(bc)'$

    The second four can be expressed as

    $ab'+ac$

    so the total expression is

    $a'(bc)'+ab'+ac$ and this minimizes to

    $(ac + a'c'+b')$
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    Re: Convert Truth Table to Boolean AND-OR Expression

    Quote Originally Posted by romsek View Post
    you certainly can just list the combinations that result in 1 in your truth table but I suspect they want you to simplify that.

    The first four entries can be seen to be

    $a'(bc)'$

    The second four can be expressed as

    $ab'+ac$

    so the total expression is

    $a'(bc)'+ab'+ac$ and this minimizes to

    $(ac + a'c'+b')$
    Can you explain how the first four entries can be seen as a'(bc)' ? What I got for the first four entries in the truth table (starting at the top) are abc + abc' + ab'c + a(bc)'. I simplified it as such:

    abc + abc' + ab'c + a(bc)' = ab(c + c') + ab'(c + c')
    = (c + c') + (ab + ab')
    = 1 + a(b + b')
    = 1 + a(1)
    = 1 + a
    = a

    I didn't try the second four since I didn't understand the first four.
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    Re: Convert Truth Table to Boolean AND-OR Expression

    Quote Originally Posted by lamentofking View Post
    Can you explain how the first four entries can be seen as a'(bc)' ? What I got for the first four entries in the truth table (starting at the top) are abc + abc' + ab'c + a(bc)'. I simplified it as such:

    abc + abc' + ab'c + a(bc)' = ab(c + c') + ab'(c + c')
    = (c + c') + (ab + ab')
    = 1 + a(b + b')
    = 1 + a(1)
    = 1 + a
    = a

    I didn't try the second four since I didn't understand the first four.
    the first four entries all correspond to a' as a is 0 for all four.

    the relationship of b and c in those first four entries is (bc)'. That should be pretty clear.

    Combine those two facts to get that the first four entries correspond to a'(bc)'

    000 - 1(00)' - 1(0)' - 11 - 1
    001 - 1(01)' - 1(0)' - 11 - 1
    010 - 1(10)' - 1(0)' - 11 - 1
    011 - 1(11)' - 1(1)' - 10 - 0
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    Re: Convert Truth Table to Boolean AND-OR Expression

    Quote Originally Posted by romsek View Post
    the first four entries all correspond to a' as a is 0 for all four.

    the relationship of b and c in those first four entries is (bc)'. That should be pretty clear.

    Combine those two facts to get that the first four entries correspond to a'(bc)'

    000 - 1(00)' - 1(0)' - 11 - 1
    001 - 1(01)' - 1(0)' - 11 - 1
    010 - 1(10)' - 1(0)' - 11 - 1
    011 - 1(11)' - 1(1)' - 10 - 0
    Sorry, my book doesn't do a good job of explaining this. Is that a rule to convert to AND-OR expression that with a value like 000 you pull out a 1 and take the complement of the last two numbers and simplify? I can do that pattern:

    000 - 1(00)' - 1(0)' - 11 - 1
    001 - 1(01)' - 1(0)' - 11 - 1
    010 - 1(10)' - 1(0)' - 11 - 1
    011 - 1(11)' - 1(1)' - 10 - 0

    for the second 4 as well?
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