Simplifica $$$\left(X \cdot Y\right) + \overline{Y \cdot Z} + \left(X \cdot \overline{Y} \cdot Z \cdot \left(\left(X \cdot Y\right) + Z\right)\right)$$$
Calculadora relacionada: Calculadora de tablas de verdad
Tu entrada
Simplifica la expresión booleana $$$\left(X \cdot Y\right) + \overline{Y \cdot Z} + \left(X \cdot \overline{Y} \cdot Z \cdot \left(\left(X \cdot Y\right) + Z\right)\right).$$$
Solución
Aplica la propiedad conmutativa:
$$\left(X \cdot Y\right) + \overline{Y \cdot Z} + \left(X \cdot \overline{Y} \cdot Z \cdot {\color{red}\left(\left(X \cdot Y\right) + Z\right)}\right) = \left(X \cdot Y\right) + \overline{Y \cdot Z} + \left(X \cdot \overline{Y} \cdot Z \cdot {\color{red}\left(Z + \left(X \cdot Y\right)\right)}\right)$$Aplica la ley de absorción $$$x \cdot \left(x + y\right) = x$$$ con $$$x = Z$$$ y $$$y = X \cdot Y$$$:
$$\left(X \cdot Y\right) + \overline{Y \cdot Z} + \left(X \cdot \overline{Y} \cdot {\color{red}\left(Z \cdot \left(Z + \left(X \cdot Y\right)\right)\right)}\right) = \left(X \cdot Y\right) + \overline{Y \cdot Z} + \left(X \cdot \overline{Y} \cdot {\color{red}\left(Z\right)}\right)$$Aplica el teorema de De Morgan $$$\overline{x \cdot y} = \overline{x} + \overline{y}$$$ con $$$x = Y$$$ y $$$y = Z$$$:
$$\left(X \cdot Y\right) + {\color{red}\left(\overline{Y \cdot Z}\right)} + \left(X \cdot \overline{Y} \cdot Z\right) = \left(X \cdot Y\right) + {\color{red}\left(\overline{Y} + \overline{Z}\right)} + \left(X \cdot \overline{Y} \cdot Z\right)$$Aplica la propiedad conmutativa:
$${\color{red}\left(\left(X \cdot Y\right) + \overline{Y} + \overline{Z} + \left(X \cdot \overline{Y} \cdot Z\right)\right)} = {\color{red}\left(\left(X \cdot Y\right) + \overline{Y} + \left(X \cdot \overline{Y} \cdot Z\right) + \overline{Z}\right)}$$Aplica la propiedad conmutativa:
$$\left(X \cdot Y\right) + \overline{Y} + {\color{red}\left(X \cdot \overline{Y} \cdot Z\right)} + \overline{Z} = \left(X \cdot Y\right) + \overline{Y} + {\color{red}\left(\overline{Y} \cdot X \cdot Z\right)} + \overline{Z}$$Aplica la ley de absorción $$$x + \left(x \cdot y\right) = x$$$ con $$$x = \overline{Y}$$$ y $$$y = X \cdot Z$$$:
$$\left(X \cdot Y\right) + {\color{red}\left(\overline{Y} + \left(\overline{Y} \cdot X \cdot Z\right)\right)} + \overline{Z} = \left(X \cdot Y\right) + {\color{red}\left(\overline{Y}\right)} + \overline{Z}$$Aplica la propiedad conmutativa:
$${\color{red}\left(\left(X \cdot Y\right) + \overline{Y} + \overline{Z}\right)} = {\color{red}\left(\overline{Y} + \left(X \cdot Y\right) + \overline{Z}\right)}$$Aplica la propiedad conmutativa:
$$\overline{Y} + {\color{red}\left(X \cdot Y\right)} + \overline{Z} = \overline{Y} + {\color{red}\left(Y \cdot X\right)} + \overline{Z}$$Aplica la ley de redundancia $$$x + \left(\overline{x} \cdot y\right) = x + y$$$ con $$$x = \overline{Y}$$$ y $$$y = X$$$:
$${\color{red}\left(\overline{Y} + \left(Y \cdot X\right)\right)} + \overline{Z} = {\color{red}\left(\overline{Y} + X\right)} + \overline{Z}$$Respuesta
$$$\left(X \cdot Y\right) + \overline{Y \cdot Z} + \left(X \cdot \overline{Y} \cdot Z \cdot \left(\left(X \cdot Y\right) + Z\right)\right) = \overline{Y} + X + \overline{Z}$$$