Prove that $x+(\overline{x}\cdot\overline{y})=x+\overline{y}$

Question

Prove that $x+(\overline{x}\cdot\overline{y})=x+\overline{y}$

The values of both these boolean functions show that these 2 are equivalent.

$x$  $\overline{x}$  $y$  $\overline{x}\cdot\overline{y}$  $x+(\overline{x}\cdot\overline{y})$  $x+\overline{y}$

0  1  0  0  0  0

0  1  1  1  1  1

1  0  0  0  1  1

1  0  1  0  1  1

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I proved this. Missed the distributive law. \begin{eqnarray*} x+(\overline{x}\cdot\overline{y})&=&(x+\overline{x})\cdot(x+\overline{y})\\ &=&1\cdot(x+\overline{y})\\ &=&x+\overline{y} \end{eqnarray*}

I often find the answer just before or after posting the question to SE. Don't know why.

Answer 1

An alternative solution to Caran's:

$x+\bar y = x+(x+\bar x)\cdot\bar y = (x+x\cdot\bar y) +\bar x\cdot \bar y = x+\bar x\cdot\bar y$

Answer 2

Using De Morgan's laws one can obtain: $$x+(\overline{x}\cdot\overline{y})=x+\overline{x+y}=\overline{\overline{x}\cdot \overline{\overline{(x+y)}}}=\overline{\overline{x}\cdot (x+y)}=\overline{\overline{x}\cdot x+\overline{x}\cdot y}=\overline{\overline{x}\cdot y}=\overline{\overline{x}}+\overline{y}=x+\overline{y}$$

Answer 3

Proved this by myself. Missed the distributive law. \begin{eqnarray*} x+(\overline{x}\cdot\overline{y})&=&(x+\overline{x})\cdot(x+\overline{y})\\ &=&1\cdot(x+\overline{y})\\ &=&x+\overline{y} \end{eqnarray*}