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Date: [20230308101739]
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![[ALGEBRA BOOLOWSKA]]



# Tautologie
## 1
$$\begin{gathered}
x\cdot y + \bar{y} \cdot z = (x+z)\cdot(y+\bar{x})
\\\\
	P=(x+z)\cdot(y+\bar{x}){=}^{10}(x+z)\cdot y + (x+z)\cdot \bar{x}\\=^{8}y\cdot(x+z)+\bar{x}\cdot(x+z)=^{10}y\cdot x+ y \cdot z + \bar{x}\cdot x + \bar{x} \cdot z\\ =^{4}y\cdot x + y \cdot z + 0 +\bar{x}\cdot z =^{3}y\cdot x + y \cdot z + \bar{x}\cdot z\\=^{2}y\cdot x + y \cdot z  \cdot 1 + \bar{x}\cdot z =^{4} y\cdot x + y\cdot z \cdot (x+\bar{x})+\bar{x}\cdot z \\=^{10}y\cdot x + y\cdot z \cdot x + y \cdot z \cdot \bar{x}+\bar{x}\cdot z =^{8} x*y+xyz+\bar{x}zy+\bar{x}z \\=^{2}xy(1+z)+\bar{x}z(y+1)=^{8}xy(z+1)+\bar{x}z(y+1)=^{2}xy\cdot1+\bar{x}z\cdot 1\\=^{2}xy+\bar{x}z = L

\end{gathered}$$





# Bramki na stykach

![[1. Algebra Boola 2023-03-08 11.11.50.excalidraw]]


## Poprzednie zadanie:
![[1. Algebra Boola 2023-03-08 11.17.37.excalidraw]]

![[1. Algebra Boola 2023-03-08 11.38.40.excalidraw]]