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smileypkr
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Posted on 09-13-07 11:54
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HI Yekloyatri here is your solution for 1 A'B +AB' + AB + A'B' = 1 rearrange them first A'B +A'B'+ AB +AB' = 1 A'(B + B') + A(B +B') = 1 (since B +B' = 1) A + A' =1 1=1 proved hope it will help do same thing with second problem
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thapap
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Posted on 09-14-07 12:07
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basic boolean rules can be referred to @ http://www.ee.surrey.ac.uk/Projects/Labview/boolalgebra/index.html#commutativelaw A'B + AB' + AB + A'B' = (A'B +A'B')+(AB'+AB) =A'(B+B')+A(B'+B) =A' (1) + A (1) = A' + A = 1 X' + XY + XZ' + XY'Z' X' + XY + XZ' (1+Y') X' + XY + XZ' X' + X(Y+Z') X' + Y + Z' [ T10 A]
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yekloyatri
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Posted on 09-14-07 12:41
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Thank you so much guys. I missed the class and i dont know anyone in my class. This is the reason i posted this stuff here. And yeah smileypkr and thapap thanks for the link. Its' great i would be digging more into it.
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