MHB Simplifying Boolean Circuits: W NAND X & YNORZ

AI Thread Summary
The discussion focuses on simplifying a Boolean expression derived from a circuit using De Morgan's theorem. The original expression is identified as f = (w NAND x) NOR (y NOR z). It is rewritten in Boolean form as f = ¬(¬(wx) + ¬(y + z)). Applying De Morgan's theorem leads to the transformation of the expression into f = wx(y + z). The conversation suggests that further simplification to Sum of Products (SoP) form may be possible, indicating a need for clarity in notation and understanding of Boolean algebra principles.
shamieh
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Find the simplified equation (SOP FORM) corresponding to the following circuit (i.e. apply Demorgan's theorem to bring the negations inside the parenthesis).

View attachment 1466So is it saying W NAND X which means (w!x!) + YNORZ so (w!x!) + (y!z!) = wxyz?
 

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shamieh said:
Find the simplified equation (SOP FORM) corresponding to the following circuit (i.e. apply Demorgan's theorem to bring the negations inside the parenthesis).

View attachment 1466So is it saying W NAND X which means (w!x!) + YNORZ so (w!x!) + (y!z!) = wxyz?

Your notation is a bit confusing to me. So let me write it as I'm used to.

You have f = (w NAND x) NOR (y NOR z).

Written in boolean form this is:
$$f = \overline{\overline{wx} + \overline{y+z}}$$
Applying De Morgan's Theorem ($\overline{p+q} = \overline p \cdot \overline q$) yields:
$$f = \overline{\overline{wx}} \cdot \overline{\overline{y+z}} = wx(y+z)$$

Perhaps you can simplify it further to SoP form?
 
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