Can anyone help with Demorgans and simpification please? The " ' " represent the nots. I wish I could write it on here better but this is the only way I see to. Some of the variables have more than one bar over them. I have tried several attempts at this and this is what I have. (C'+D')A''C'''D''(A''C'+D'')=X one bar is over the entire equation C'+D' + A''C'''D'' + A''C'+D'' C'+D' + AC'D + (A''+C')(D'') C'+D' + AC'D + AD + C'D C'+D' + C'D(A+1) + AD C'+D' + C'D + AD C'D' + C'D + AD C'(D'+D) + AD C' +AD Thanks
jared11378, The first thing to do is to get your notation straight. No one can solve your problem until they know what the expression is. For instance, one of the variables is C'''. Does this mean negation 3 times which is C' ? I will write it the way I am guessing it is so you can see how it should be written. No doubt that will result in a different espresson from what you intended. There should never be more than one "not" unless there is a parenthesis present. And the number of left parenthesis should equal the number of right parentheses. Ratch (((((C'+D')A')'C')'D')'(A'C'+D')')'
I wish I had a scanner but I dont or this would be easier. Yes the C''' does eventually reduce to C'. To start the problem though there is 1 bar over the whole expression the way I have it written. Every time you see a negation " ' " add a bar over that variable or equation if all the variables have a second negation. One of the bars is over the entire expression though. I think this is how you are wanting me to write it but I'm not 100% sure. (((((C'+D')A')'C')D')'(A'C+D')') I hope this helps.
jared11378, The outer parantheses is without a "not", so you might as well remove it. Is that what you want? Do you really understand what the parenthesis are doing? A parenthesis with a "not" is the same as a bar over the expression the parenthesis encompasses. Ratch
The first thing I recommend you do is multiply out the brackets, inparticular this part: ((((C'+D')A')'C')D')' You can then start with by double-NOTing the function and performing a two-part dot-DeMorganisation, i.e. (A.B)' = (A' + B') Dave
OK here is what I get so far. Its confusing because I can work it differantely and get another answer. I know there only one right way but Im having trouble figuring out which path is right to take. How do you know? ((((C'+D')A')'C')D')'(A'C+D')' (C'+D') + A''C'''D'' + (A''C'+D'') (C'+D') + A''C'''D'' + (A''C'D'') (C'+D') + AC'D + AC'D (C'+D') + 1 (C'+D')
The first step doesn't make sense. Before you can Demorganise an expression you need to remove the embedded brackets (make sure the brackets are correct as written here because they are very important to the structure of the expression). Remember Boolean Algebra is distributive: ((((C'+D')A')'C')D')'(A'C+D')' Becomes (((A'C'+A'D')'C')D')'(A'C+D')' Two-part Demoganise on (A'C'+A'D')' gives (A'C')'(A'D')' So you now have: (((A'C')'(A'D')'C')D')'(A'C+D')' DeMorganise (A'C')' = (A'' + C'') = (A + C) DeMorganise (A'D')' = (A'' + D'') = (A + D) So you now have: (((A + C)(A + D)C')D')'(A'C+D')' Now multiply out ((A + C)(A + D)C') And so on. Make sense? Write it out if it is unclear. You need to divide and conquer to remove the embedded brackets otherwise it is unworkable. Just check my work about I have done this in my head on screen so there may be a stray bracket or NOT-function. Dave
jared11378, Start from the outside and work toward the inside. ((((C'+D')A')'C')D')'(A'C+D')' = ((((C'+D')A')'C')' + D )(A'C)'D = etc You can probably work out the rest. Ratch
I attatched a copy of the problem exactly how it shows. I just want to make sure were on the same track
jared11378 I guess we are not. The expression from your pdf file would be ((C + D)(AC'D)'(A'C + D'))' . If that is the problem, work from the outside toward the inside. Get rid of the outermost bar first, then work on each inner term that has a bar. Ratch
Your expression is considerably easier than you wrote up - there are no embedded brackets to remove before DeMorganising the function. I'm with Ratch on the approach. Might be an idea to drop a bracket around the AC'D like Ratch has done to make it easier to understand. Start by applying (ABC)' = (A' + B' + C') to the whole expression. And Remember you can double-NOT the function without changing the expression: A = A'' Dave
((C + D)(AC'D)'(A'C + D'))' (C+D)' + ((AC'D)')' + (A'C + D') (C+D) + ((AC'D)')' + (AC'D) (C+D) + AC'D + (AC'D) (C+D) + (AC'D + AC'D)=1 C+D Would this be it? I can get different answers if I turn (A'C + D'))' into (A+C')D. How do you know what to Demorgan and what to leave the way it is?
You have miswrote this. Should be: (C+D)' + ((AC'D)')' + (A'C + D')' This step is wrong. Should be: (C+D)' + ((AC'D)')' + (A'C + D')' (C+D)' + (AC'D) + (A'C + D')' Then Demorganise (C+D)' and (A'C + D')' So you get: (C'D') + (AC'D) + ((A'C)'D'') Which simplifies to: (C'D') + (AC'D) + ((A'C)'D) Then Demorganise (A'C)' So you get: (C'D') + (AC'D) + ((A'' + C')D) Which simplifies to: (C'D') + (AC'D) + ((A + C')D) Multiply out the brackets and look to simplify the expression. Note you need to know the identity (A + AB) = A. I have just done a quick work through and I get the answer C' +AD Practice and experience. Dave
I see it now! ((C+D)' + (((AC'D)')' + (A'C + D')' ((C+D)' + (AC'D) + (A'C + D')' (C'D') + (AC'D) + (A'C)'D'') (C'D') + (AC'D) + ((A''+C')D) (C'D') + (AC'D) + ((A+C')D) (C'D') + (AC'D) + AD+C'D C'D' + C'D(A+1) + AD C'D' + C'D + AD C'(D' + D) + AD C' + AD It works with the truth table. I just dont understand how I can get so many differant answers. Thanks for the help though.
jared11378, No you don't. Do remember in post #2 of this thread I said, "And the number of left parentheses should equal the number of right parentheses." You have 6 left parentheses and 4 right parentheses. That is a red flag indicating something is wrong in how you have defined the expression. Ratch
Ratch, you are right about the number of parenthesis, however if you follow jared's working-out you will see that he has done the algebra correctly. The erroneous parenthesis do not change the fundamental nature of the expression (i.e. there are not embedded brackets where there shouldn't be as was the case previously). The parenthesis anomaly corrects itself in line 4 of his working-out. Dave
Agreed. I would be tempted to put it down to the fact that writing up long expression onto the forums is potentially error prone. I hope jared had it written correctly on paper when it was submitted otherwise a few marks may go missing. Dave