## Exercises 5.2 Exercises

###### 1.

Prove the identity property expressed by Equation (5.1.3) and Equation (5.1.4).

For Equation (5.1.3):

\(x\) | \(x \cdot 1\) |

\(\binary{0}\) | \(\binary{0}\) |

\(\binary{1}\) | \(\binary{1}\) |

And for Equation (5.1.4):

\(x\) | \(x + 0\) |

\(\binary{0}\) | \(\binary{0}\) |

\(\binary{1}\) | \(\binary{1}\) |

###### 2.

Prove the commutative property expressed by Equation (5.1.5) and Equation (5.1.6).

For Equation (5.1.5):

\(x\) | \(y\) | \(x \cdot y\) | \(y \cdot x\) |

\(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |

\(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) |

\(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |

\(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |

And for Equation (5.1.6):

\(x\) | \(y\) | \(x + y\) | \(y + x\) |

\(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |

\(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |

\(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) |

\(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |

###### 3.

Prove the annulment property expressed by Equation (5.1.7) and Equation (5.1.8).

For Equation (5.1.7):

\(x\) | \(x \cdot 0\) |

\(\binary{0}\) | \(\binary{0}\) |

\(\binary{1}\) | \(\binary{0}\) |

And for Equation (5.1.8):

\(x\) | \(x + 1\) |

\(\binary{0}\) | \(\binary{1}\) |

\(\binary{1}\) | \(\binary{1}\) |

###### 4.

Prove the complement property expressed by Equation (5.1.9) and Equation (5.1.10).

For Equation (5.1.9):

\(x\) | \(x'\) | \(x \cdot x'\) |

\(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) |

\(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) |

And for Equation (5.1.10):

\(x\) | \(x'\) | \(x + x'\) |

\(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) |

\(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) |

###### 5.

Prove the idempotent property expressed by Equation (5.1.11) and Equation (5.1.12).

For Equation (5.1.11):

\(x\) | \(x\) | \(x \cdot x\) |

\(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |

\(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |

And for Equation (5.1.12):

\(x\) | \(x\) | \(x + x\) |

\(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |

\(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |

###### 6.

Prove the distributive property expressed by Equation (5.1.13) and Equation (5.1.14).

For Equation (5.1.13):

\(x\) | \(y\) | \(z\) | \(y + z\) | \(x \cdot (y + z)\) | \(x \cdot y\) | \(x \cdot z\) | \(x \cdot y + x \cdot z\) |

\(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |

\(\binary{0}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |

\(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |

\(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |

\(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |

\(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) |

\(\binary{1}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) |

\(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |

And for Equation (5.1.14):

\(x\) | \(y\) | \(z\) | \(y \cdot z\) | \(x + y \cdot z\) | \(x + y\) | \(x + z\) | \((x + y) \cdot (x + z)\) |

\(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |

\(\binary{0}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) |

\(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) |

\(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |

\(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |

\(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |

\(\binary{1}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |

\(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |