Boolean Identities
The truth tables for the identity $x+yz=(x+y)(x+z)$ are:
| $x$ | $y$ | $z$ | $yz$ | $\boldsymbol{x + yz}$ |
|---|---|---|---|---|
| $0$ | $0$ | $0$ | $0$ | $0$ |
| $0$ | $0$ | $1$ | $0$ | $0$ |
| $0$ | $1$ | $0$ | $0$ | $0$ |
| $0$ | $1$ | $1$ | $1$ | $1$ |
| $1$ | $0$ | $0$ | $0$ | $1$ |
| $1$ | $0$ | $1$ | $0$ | $1$ |
| $1$ | $1$ | $0$ | $0$ | $1$ |
| $1$ | $1$ | $1$ | $1$ | $1$ |
| $x$ | $y$ | $z$ | $x + y$ | $x + z$ | $\boldsymbol{(x+y)(x+z)}$ |
|---|---|---|---|---|---|
| $0$ | $0$ | $0$ | $0$ | $0$ | $0$ |
| $0$ | $0$ | $1$ | $0$ | $1$ | $0$ |
| $0$ | $1$ | $0$ | $1$ | $0$ | $0$ |
| $0$ | $1$ | $1$ | $1$ | $1$ | $1$ |
| $1$ | $0$ | $0$ | $1$ | $1$ | $1$ |
| $1$ | $0$ | $1$ | $1$ | $1$ | $1$ |
| $1$ | $1$ | $0$ | $1$ | $1$ | $1$ |
| $1$ | $1$ | $1$ | $1$ | $1$ | $1$ |
Because the sequences of bits that we got under the $x + yz$ and $(x+y)(x+z)$ columns are exactly the same: $[0, 0, 0, 1, 1, 1, 1, 1]$, this means that the expressions are equivalent, so $x + yz = (x+y)(x+z)$ is indeed a valid identity!
These notes by Miriam Briskman are licensed under CC BY-NC 4.0 and based on sources.