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I'm trying to simplify the following expression:

(x+ ~x)~(~(~z xor y) + (yx))
   (1)~(~(~z xor y) + (yx))
   ~(~(~z xor y) + (yx))
   (~z xor y) + ~(yx)
   (~z xor y) + (~y + ~x)
   (~y~z + yz) + (~y + ~x) <---- im stuck here

How can I simplify this more? and once it is simplified how do i calculate the number of gate delays

Cesa Salaam
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  • Following FabioSomenzi's answer, you get $x' + yz + y' + y'z'$ which is equal to $x' +yz + y'z'$, and I'm not seeing any further simplification here... – amrsa Nov 22 '16 at 10:19
  • @amrsa Check your use of absorption. – Fabio Somenzi Nov 22 '16 at 17:50
  • @FabioSomenzi Right, I meant is equal to $x' + yz + y'$ and this in turn equals $x' + y' + z$ (because, in general, $a+a'b = a + b$). Thanks for the comment :) – amrsa Nov 22 '16 at 17:57

1 Answers1

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Observe that $+$ is both associative and commutative, which allows you to drop parentheses and rearrange terms.

Fabio Somenzi
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