ketz
f=x'y'z+(x'y'z)'. simplify the boolean expression. urgent



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hartnn
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you use the identity,
A+A' = 1
so, f=x'y'z+(x'y'z)' = 1 (where x'y'z can be considered as A)

ketz
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but if i proceed by expanding the bracket! I'm not getting the answer!!!!

hartnn
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expanding the bracket ?
like x'y'z +xyz' ?
no need for that, you can use A=x'y'z and that A+A'=1
you can also verify it using truth table.

ketz
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yeh i know but i want to proceed by expanding the bracket using de morgan's laws. and am not getting the answer. can anyone try it by this method!!

hartnn
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let me try
on, left of +
x'y' = (x+y)' > x'y'z =(x+y)'z
on right of +
(x'y'z)' = (x+y+z')
hmm

ganeshie8
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f=x'y'z+(x'y'z)'
x'y'z + x + y + z'
y'z + x + y + z'
x + y'z + y + z'
x + y + z + z'
x + y + 1
1

ganeshie8
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in 2ns line use this rule : x+x'y = x+y

ketz
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Can you prove the law x+x'y=x+y???

ketz
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By Boolean Algebra!

ganeshie8
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did u try to prove already... it wont be difficult i think

ketz
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No

ketz
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How can you prove or simplify it by boolean algebra??

ganeshie8
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do u knw distributive property for AND logic ?

ganeshie8
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x + (y.z) = (x+y).(x+z)

ganeshie8
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u familiar wid that prop ? u can use that to prove x + x'y = x + y very easily

ketz
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ok got it!

ganeshie8
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great :)