# Trying to simplify the following boolean expression [closed]

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:

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

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

• Hi, I am confused by what you've written. Can you please demonstrate step by step? What have I done wrong? Commented Mar 7 at 17:10
• 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 $y$. Commented Mar 7 at 17:37

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. $$y \quad \textit{Absorption Axiom}$$
• I believe step 3 should be $y(x+y)$; otherwise 100%. Commented Mar 9 at 4:54
• @mjqxxxx what is "100% otherwise"? Commented Mar 9 at 14:31
• Other than that small typo, you are 100% correct. Commented Mar 9 at 21:26
• @mjqxxxx, this is wrong. Why are you misleading me? Commented Mar 10 at 17:10