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Simplify:

$y \times [x + (x' \times y)]$

My attempt:

  1. $y \times [(x + x') \times (x + y)] \quad \textit{First Distributive Axiom} $
  2. $y \times [1 \times (x + y)] \quad \textit{First Inverse Axiom} $

And for the continued steps, I am unfortunately stuck again.

Axioms available:

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2 Answers 2

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I don't see $yy=y$ directly in the axioms you have, but it follows from $yy=y(y+0)=y$. And so

$$ y(x + x'y)=yx+yx'y=yx+yyx'=yx+yx'=y(x+x')=y(1)=y. $$

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  • $\begingroup$ Hi, I am confused by what you've written. Can you please demonstrate step by step? What have I done wrong? $\endgroup$ Commented Mar 7 at 17:10
  • $\begingroup$ What you did is fine… possibly even more clear than my answer. You can apply Identity next, to get $y(x+y)$, and finally Absorption to get $y$. $\endgroup$
    – mjqxxxx
    Commented Mar 7 at 17:37
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Solution:

  1. $y \times [(x + x') \times (x + y)] \quad \textit{First Distributive Axiom} $
  2. $y \times [1 \times (x + y)] \quad \textit{First Inverse Axiom} $
  3. $y(x+y) \quad \textit{Identity Axiom} $
  4. $yx + yy \quad \textit{Distributive Axiom} $
  5. $(yx) + y \quad \textit{Idempotent Axiom} $
  6. $y + (yx) \quad \textit{Commutative Axiom} $
  7. $y \quad \textit{Absorption Axiom} $
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  • $\begingroup$ I believe step 3 should be $y(x+y)$; otherwise 100%. $\endgroup$
    – mjqxxxx
    Commented Mar 9 at 4:54
  • $\begingroup$ @mjqxxxx what is "100% otherwise"? $\endgroup$ Commented Mar 9 at 14:31
  • $\begingroup$ Other than that small typo, you are 100% correct. $\endgroup$
    – mjqxxxx
    Commented Mar 9 at 21:26
  • $\begingroup$ @mjqxxxx, this is wrong. Why are you misleading me? $\endgroup$ Commented Mar 10 at 17:10

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