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Boolean Algebra help.?
(x + y).(x' + z) = x.z + x'y Its asking the name of the theorem and how to solve using the first 8 laws of Boolean Algebra. I know that law 16 states that (x+y) (x'+z) = xz + x'y so this proves this to be true but I"m not entirely sure on how to calculate using the first 8 laws. Here's what I got so far: xx' + xz + x'y + yz == 0 + x'y + z(x + y), from here I'm not sure where to go.
2 Answers
- 1 decade agoFavourite answer
You could split your proof into two cases: yz=1 and yz=0
What you have so far completes the yz=0 case.
Then yz=1 ==> y=1 and z=1
Hence (x+y)(x'+z)=(x+1)(x'+1)=(1)(1)=1
xz+x'y = x(1)+x'(1)=x+x'=1
- 5 years ago
God is above us, numbers lower than us, life questions aside. concept includes some secret, numbers are by potential of definition and under no circumstances something hidden. yet maximum of all, there is not any concept in any respect in Boolean Algebra, both contained in the elements or contained in the operations. that is not any longer a logical fallacy. i'm particular you ought to throw mutually a syllogism of unimpeachable form , besides the indisputable fact that it would resemble All cows are strong ballet dancers Bessie is a cow Bessie is a robust ballet dancer. that is merely unfaithful.
