simplify(~(~(x | y) & ~(y | z)))

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Lisa Lee
Lisa Lee 2017-8-10
编辑： Jan 2017-8-10
I don't understand why it is equivalent to x | y | z . Can someone please help to explain this to me?

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Jan
Jan 2017-8-10
编辑：Jan 2017-8-10
Remember these equivalences ( De Morgan's laws ):
~(x | y) <==> ~x & ~y
~(x & y) <==> ~x | ~y
Then
~(~(x | y) & ~(y | z)) ==>
~((~x & ~y) & (~y & ~z)) ==>
~(~x & ~y) | ~(~y & ~z) ==>
~~x | ~~y | ~~y | ~~z ==>
x | y | y | z ==>
x | y | z
Applying the operations from the outside to the inside is faster:
~(~(x | y) & ~(y | z)) ==>
~~(x | y) | ~~(x | z) ==>
x | y | y | z

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