Portas logicas

Teorema de álgebra booleana
Exercício
A) A+AB=A A | B | AB | S | 0 | 0 | | | 0 | 1 | | | 1 | 0 | | | 1 | 1 | | |

B) A . (A+B)=A A | B | A+B | S | 0 | 0 | | | 0 | 1 | | | 1 | 0 | | | 1 | 1 | | |

C) AB+AB̅=A A | B | B̅ | A.B | A.B̅ | S | 0 | 0 | 1 | | | | 0 | 1 | 0 | | | | 1 | 0 | 1 | | | | 1 | 1 | 0 | | | |

D) (A+B) . (A+B̅)=A A | B | B̅ | A+B | A+B̅ | S | 0 | 0 | 1 | | | | 0 | 1 | 0 | | | | 1 | 0 | 1 | | | | 1 | 1 | 0 | | | |

E) A+A̅B = A+B A | B | A̅ | A̅ .B | S | 0 | 0 | 1 | | | 0 | 1 | 1 | | | 1 | 0 | 0 | | | 1 | 1 | 0 | | | A | B | S | 0 | 0 | | 0 | 1 | | 1 | 0 | | 1 | 1 | |

F) A . (A̅+B) = AB A | B | A̅ | A̅ +B | S | 0 | 0 | 1 | | | 0 | 1 | 1 | | | 1 | 0 | 0 | | | 1 | 1 | 0 | | | A | B | S | 0 | 0 | | 0 | 1 | | 1 | 0 | | 1 | 1 | |

G) A+BC = (A+B) . (A+C) A | B | C | B . C | S | 0 | 0 | 0 | | | 0 | 0 | 1 | | | 0 | 1 | 0 | | | 0 | 1 | 1 | | | 1 | 0 | 0 | | | 1 | 0 | 1 | | | 1 | 1 | 0 | | | 1 | 1 | 1 | | | A | B | C | A+B | A+C | S | 0 | 0 | 0 | | | | 0 | 0 | 1 | | | | 0 | 1 | 0 | | | | 0 | 1 | 1 | | | | 1 | 0 | 0 | | | | 1 | 0 | 1 | | | | 1 | 1 | 0 | | | | 1 | 1 | 1 | | | |

H) A.(B+C) = AB + AC A | B | C | B+C | S | 0 | 0 | 0 | | | 0 | 0 | 1 | | | 0 | 1 | 0 | | | 0 | 1 | 1 | | | 1 | 0 | 0 | | | 1 | 0 | 1 | | | 1 | 1 | 0 | | | 1 | 1 | 1 | | | A | B | C | A . B | A . C | S | 0 | 0 | 0 | | | | 0 | 0 | 1 | | | | 0 | 1 | 0 | | | | 0 | 1 | 1 | | | | 1 | 0 | 0 | | | | 1 | 0 | 1 | | | | 1 | 1 | 0 | | | | 1 | 1 | 1 | | | |

I) AB+A̅C = (A+B) . (A̅+B) A | B | C | A̅ | A . B | A̅ . C | S | 0 | 0 | 0 | 1 | | | | 0 | 0 | 1 | 1 | | | | 0 | 1 | 0 | 1 | | |