# Portas logicas

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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 | |...